{"answer": "6.247", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF014", "CMF011"], "effective_parameters": {"Biom_plant_yesterday": 704.9, "Coef_Rm": 0.0039, "DVI": 0.48, "Tmn": 24.8}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless.\nThe computed value is 6.247 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{6.247}.\n#### 6.247", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "wrong_formula_id": "CMF014"} {"answer": "72.07", "benchmark_version": "cropmath-v1", "change": {"new_value": 30.0, "old_value": 29.3, "parameter": "To"}, "distractor_formula_ids": ["CMF033", "CMF037"], "effective_parameters": {"ABIOMASS": 13378.0, "PDT": 7.0, "Q10": 2.18, "T": 24.3, "To": 30.0}, "family": "growth_yield", "formula": {"expression": "(0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)", "id": "CMF048", "latex": "RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}", "name": "维持呼吸", "name_en": "Maintenance respiration", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF048_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]", "knowledge_formula_expr": "Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Product with Q10 temperature response", "mode": "sensitivity", "parameters": {"ABIOMASS": 13378.0, "PDT": 7.0, "Q10": 2.18, "T": 24.3, "To": 29.3}, "problem": "In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration.\nUse the verified formula:\n\\[RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\\]\nIn calculation-friendly notation, this is:\nRM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10).\nSubstitute the relevant values:\nPDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=30.0 C.\nThe computed value is 72.07 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{72.07}.\n#### 72.07", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF033"} {"answer": "28.34", "benchmark_version": "cropmath-v1", "change": {"new_value": 31.5, "old_value": 28.6, "parameter": "Tair"}, "distractor_formula_ids": ["CMF025", "CMF023"], "effective_parameters": {"Tair": 31.5}, "family": "methane_emission", "formula": {"expression": "4.4 + 0.76 * Tair", "id": "CMF024", "latex": "T_{soil}=4.4+0.76T_{air}", "name": "土壤温度估算", "name_en": "Soil temperature from air temperature", "precision": 2, "unit": "C"}, "id": "CMF024_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]", "knowledge_formula_expr": "Formula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_name": "Relevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Linear", "mode": "sensitivity", "parameters": {"Tair": 28.6}, "problem": "Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil temperature from air temperature.\nUse the verified formula:\n\\[T_{soil}=4.4+0.76T_{air}\\]\nIn calculation-friendly notation, this is:\nTsoil = 4.4 + 0.76 * Tair.\nSubstitute the relevant values:\nTair=31.5 C.\nThe computed value is 28.34 C.\nTherefore, the final numeric answer is \\boxed{28.34}.\n#### 28.34", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF025"} {"answer": "0.708", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF024", "CMF023"], "effective_parameters": {"WaterC": 0.606}, "family": "methane_emission", "formula": {"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)", "id": "CMF025", "latex": "WI=0.49e^{3.88W_c-5.4W_c^2}", "name": "水分分解因子", "name_en": "Water-content decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF025_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]", "knowledge_formula_expr": "Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "parameters": {"WaterC": 0.606}, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Water-content decomposition factor.\nUse the verified formula:\n\\[WI=0.49e^{3.88W_c-5.4W_c^2}\\]\nIn calculation-friendly notation, this is:\nWI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).\nSubstitute the relevant values:\nWaterC=0.606 fraction.\nThe computed value is 0.708 fraction.\nTherefore, the final numeric answer is \\boxed{0.708}.\n#### 0.708", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF024"} {"answer": "50.61", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF002", "CMF010"], "effective_parameters": {"GrainYield": 8113.0, "NUptake": 160.3}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "GrainYield / NUptake", "id": "CMF003", "latex": "NUE=\\frac{GrainYield}{NUptake}", "name": "氮利用效率", "name_en": "Nitrogen use efficiency", "precision": 2, "unit": "kg grain/kg N"}, "id": "CMF003_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]", "knowledge_formula_expr": "Formula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_name": "Relevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Ratio", "mode": "direct", "parameters": {"GrainYield": 8113.0, "NUptake": 160.3}, "problem": "Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Nitrogen use efficiency.\nUse the verified formula:\n\\[NUE=\\frac{GrainYield}{NUptake}\\]\nIn calculation-friendly notation, this is:\nNUE = GrainYield / NUptake.\nSubstitute the relevant values:\nGrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha.\nThe computed value is 50.61 kg grain/kg N.\nTherefore, the final numeric answer is \\boxed{50.61}.\n#### 50.61", "split": "dev", "subject": "Nitrogen", "template_id": 1, "wrong_formula_id": "CMF002"} {"answer": "1.04", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.59, "old_value": 0.65, "parameter": "beta"}, "distractor_formula_ids": ["CMF030", "CMF034"], "effective_parameters": {"Cx": 363.0, "beta": 0.59}, "family": "growth_yield", "formula": {"expression": "1 + beta * ln(Cx / 340)", "id": "CMF029", "latex": "F_{CO2}=1+\\beta\\ln(C_x/340)", "name": "二氧化碳影响因子", "name_en": "CO2 response factor", "precision": 2, "unit": "multiplier"}, "id": "CMF029_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]", "knowledge_formula_expr": "Formula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "parameters": {"Cx": 363.0, "beta": 0.65}, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as CO2 response factor.\nUse the verified formula:\n\\[F_{CO2}=1+\\beta\\ln(C_x/340)\\]\nIn calculation-friendly notation, this is:\nFCO2 = 1 + beta * ln(Cx / 340).\nSubstitute the relevant values:\nCx=363.0 ppm, beta=0.59 dimensionless.\nThe computed value is 1.04 fraction.\nTherefore, the final numeric answer is \\boxed{1.04}.\n#### 1.04", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF030"} {"answer": "183.14", "benchmark_version": "cropmath-v1", "change": {"new_value": 223.6, "old_value": 248.4, "parameter": "SLAc"}, "distractor_formula_ids": ["CMF051", "CMF044"], "effective_parameters": {"GDD": 611.2, "SLAc": 223.6}, "family": "growth_yield", "formula": {"expression": "if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)", "id": "CMF053", "latex": "SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)", "name": "比叶面积", "name_en": "Specific leaf area", "precision": 2, "unit": "cm2/g"}, "id": "CMF053_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]", "knowledge_formula_expr": "Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_name": "Relevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "parameters": {"GDD": 611.2, "SLAc": 248.4}, "problem": "In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Specific leaf area.\nUse the verified formula:\n\\[SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\\]\nIn calculation-friendly notation, this is:\nSLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200).\nSubstitute the relevant values:\nGDD=611.2 C day, SLAc=223.6 cm2/g.\nThe computed value is 183.14 cm2/g.\nTherefore, the final numeric answer is \\boxed{183.14}.\n#### 183.14", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF051"} {"answer": "0.35", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF036", "CMF053"], "effective_parameters": {"PDT": 47.6, "a": 0.053}, "family": "growth_yield", "formula": {"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))", "id": "CMF030", "latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)", "name": "生理年龄因子", "name_en": "Physiological age factor", "precision": 2, "unit": "fraction"}, "id": "CMF030_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]", "knowledge_formula_expr": "Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "parameters": {"PDT": 47.6, "a": 0.053}, "problem": "A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Physiological age factor.\nUse the verified formula:\n\\[F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\\]\nIn calculation-friendly notation, this is:\nFPA = if PDT<28: 1; else: exp(-a*(PDT-28)).\nSubstitute the relevant values:\nPDT=47.6 physiological days, a=0.053 dimensionless.\nThe computed value is 0.35 fraction.\nTherefore, the final numeric answer is \\boxed{0.35}.\n#### 0.35", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF036"} {"answer": "56.2", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF048", "CMF042"], "effective_parameters": {"SDEPTH": 1.6}, "family": "growth_yield", "formula": {"expression": "45 + 7 * SDEPTH", "id": "CMF028", "latex": "EMGDD=45+7SDEPTH", "name": "出苗热量需求", "name_en": "Emergence thermal requirement", "precision": 1, "unit": "C day"}, "id": "CMF028_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]", "knowledge_formula_expr": "Formula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear", "mode": "direct", "parameters": {"SDEPTH": 1.6}, "problem": "In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Emergence thermal requirement.\nUse the verified formula:\n\\[EMGDD=45+7SDEPTH\\]\nIn calculation-friendly notation, this is:\nEMGDD = 45 + 7 * SDEPTH.\nSubstitute the relevant values:\nSDEPTH=1.6 cm.\nThe computed value is 56.2 C day.\nTherefore, the final numeric answer is \\boxed{56.2}.\n#### 56.2", "split": "dev", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF048"} {"answer": "0.09", "benchmark_version": "cropmath-v1", "change": {"new_value": 44.7, "old_value": 38.9, "parameter": "PDT"}, "distractor_formula_ids": ["CMF032", "CMF038"], "effective_parameters": {"PDT": 44.7}, "family": "growth_yield", "formula": {"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)", "id": "CMF043", "latex": "PIRO=1-PISH", "name": "地下部分配指数", "name_en": "Root partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF043_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root partitioning index.\nUse the verified formula:\n\\[PIRO=1-PISH\\]\nIn calculation-friendly notation, this is:\nPIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).\nSubstitute the relevant values:\nPDT=44.7 physiological days.\nThe computed value is 0.09 fraction.\nTherefore, the final numeric answer is \\boxed{0.09}.\n#### 0.09", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF032"} {"answer": "0.241", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF045", "CMF035"], "effective_parameters": {"PDT": 2.2}, "family": "growth_yield", "formula": {"expression": "0.0087 * PDT + 0.2222", "id": "CMF036", "latex": "K=0.0087PDT+0.2222", "name": "冠层消光系数", "name_en": "Canopy extinction coefficient", "precision": 3, "unit": "dimensionless"}, "id": "CMF036_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless", "knowledge_name": "Relevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear", "mode": "direct", "parameters": {"PDT": 2.2}, "problem": "In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Canopy extinction coefficient.\nUse the verified formula:\n\\[K=0.0087PDT+0.2222\\]\nIn calculation-friendly notation, this is:\nK = 0.0087 * PDT + 0.2222.\nSubstitute the relevant values:\nPDT=2.2 physiological days.\nThe computed value is 0.241 dimensionless.\nTherefore, the final numeric answer is \\boxed{0.241}.\n#### 0.241", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF045"} {"answer": "0.763", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF015", "CMF017"], "effective_parameters": {"Com": 3.488, "Cr": 4.31, "FEh": 0.428, "TI": 0.722}, "family": "methane_emission", "formula": {"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))", "id": "CMF022", "latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))", "name": "甲烷产生量", "name_en": "Methane production", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF022_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]", "knowledge_formula_expr": "Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "parameters": {"Com": 3.488, "Cr": 4.31, "FEh": 0.428, "TI": 0.722}, "problem": "A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane production.\nUse the verified formula:\n\\[P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\\]\nIn calculation-friendly notation, this is:\nP = max(0, 0.27 * FEh * (TI * Cr + Com)).\nSubstitute the relevant values:\nFEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day.\nThe computed value is 0.763 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.763}.\n#### 0.763", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF015"} {"answer": "0.993", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF007", "CMF011"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.428, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop water stress factor.\nUse the verified formula:\n\\[F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\\]\nIn calculation-friendly notation, this is:\nF_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily dry matter increment.\nUse the verified formula:\n\\[GCR=\\frac{DTGA-RM-RG}{1-b}\\]\nIn calculation-friendly notation, this is:\nGCR = (DTGA - RM - RG) / (1 - b).\nSubstitute the relevant values:\nDTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction.\nThe computed value is 244.63 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{244.63}.\n#### 244.63", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF041"} {"answer": "1.30", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.55, "old_value": 0.5, "parameter": "beta"}, "distractor_formula_ids": ["CMF037", "CMF043"], "effective_parameters": {"Cx": 583.0, "beta": 0.55}, "family": "growth_yield", "formula": {"expression": "1 + beta * ln(Cx / 340)", "id": "CMF029", "latex": "F_{CO2}=1+\\beta\\ln(C_x/340)", "name": "二氧化碳影响因子", "name_en": "CO2 response factor", "precision": 2, "unit": "multiplier"}, "id": "CMF029_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]", "knowledge_formula_expr": "Formula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "parameters": {"Cx": 583.0, "beta": 0.5}, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as CO2 response factor.\nUse the verified formula:\n\\[F_{CO2}=1+\\beta\\ln(C_x/340)\\]\nIn calculation-friendly notation, this is:\nFCO2 = 1 + beta * ln(Cx / 340).\nSubstitute the relevant values:\nCx=583.0 ppm, beta=0.55 dimensionless.\nThe computed value is 1.30 fraction.\nTherefore, the final numeric answer is \\boxed{1.30}.\n#### 1.30", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF037"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF034", "CMF030"], "effective_parameters": {"GDD": 203.4, "LAI0": 0.279, "RP": 0.0063}, "family": "growth_yield", "formula": {"expression": "LAI0 * exp(RP * GDD)", "id": "CMF037", "latex": "LAI_P=LAI_0e^{RP\\cdot GDD}", "name": "潜在叶面积指数", "name_en": "Potential leaf area index", "precision": 2, "unit": "m2/m2"}, "id": "CMF037_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]", "knowledge_formula_expr": "Formula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_name": "Relevant formula name: Potential leaf area index (潜在叶面积指数).\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Exponential growth", "mode": "direct", "parameters": {"GDD": 203.4, "LAI0": 0.279, "RP": 0.0063}, "problem": "Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Potential leaf area index (潜在叶面积指数).\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Potential leaf area index.\nUse the verified formula:\n\\[LAI_P=LAI_0e^{RP\\cdot GDD}\\]\nIn calculation-friendly notation, this is:\nLAIP = LAI0 * exp(RP * GDD).\nSubstitute the relevant values:\nLAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day.\nThe computed value is 1.00 m2/m2.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF034"} {"answer": "12.52", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF053", "CMF038"], "effective_parameters": {"PS": 0.246, "Po": 10.5}, "family": "growth_yield", "formula": {"expression": "Po + sqrt(1 / PS)", "id": "CMF040", "latex": "P_c=P_o+\\sqrt{1/PS}", "name": "临界日长", "name_en": "Critical daylength", "precision": 2, "unit": "h"}, "id": "CMF040_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]", "knowledge_formula_expr": "Formula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_name": "Relevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Power response", "mode": "direct", "parameters": {"PS": 0.246, "Po": 10.5}, "problem": "A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Critical daylength.\nUse the verified formula:\n\\[P_c=P_o+\\sqrt{1/PS}\\]\nIn calculation-friendly notation, this is:\nPc = Po + sqrt(1 / PS).\nSubstitute the relevant values:\nPo=10.5 h, PS=0.246 dimensionless.\nThe computed value is 12.52 h.\nTherefore, the final numeric answer is \\boxed{12.52}.\n#### 12.52", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF053"} {"answer": "1.222", "benchmark_version": "cropmath-v1", "change": {"new_value": 551.0, "old_value": 501.0, "parameter": "CO2"}, "distractor_formula_ids": ["CMF014", "CMF008"], "effective_parameters": {"CO2": 551.0, "Coef_B_CO2": 0.46}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "1 + Coef_B_CO2 * ln(CO2 / 340)", "id": "CMF001", "latex": "F_{CO2}=1+B\\ln(CO2/340)", "name": "作物二氧化碳施肥因子", "name_en": "carbon-nitrogen cycling CO2 fertilization factor", "precision": 3, "unit": "multiplier"}, "id": "CMF001_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]", "knowledge_formula_expr": "Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling CO2 fertilization factor.\nUse the verified formula:\n\\[F_{CO2}=1+B\\ln(CO2/340)\\]\nIn calculation-friendly notation, this is:\nF_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340).\nSubstitute the relevant values:\nCO2=551.0 ppm, Coef_B_CO2=0.46 dimensionless.\nThe computed value is 1.222 fraction.\nTherefore, the final numeric answer is \\boxed{1.222}.\n#### 1.222", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF014"} {"answer": "0.802", "benchmark_version": "cropmath-v1", "change": {"new_value": 33.3, "old_value": 29.0, "parameter": "Td"}, "distractor_formula_ids": ["CMF008", "CMF010"], "effective_parameters": {"Coef_T_M": 0.13, "Td": 33.3, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)", "id": "CMF013", "latex": "F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}", "name": "作物温度活性因子", "name_en": "carbon-nitrogen cycling crop temperature activity factor", "precision": 3, "unit": "response factor"}, "id": "CMF013_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]", "knowledge_formula_expr": "Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "parameters": {"Coef_T_M": 0.13, "Td": 29.0, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "problem": "In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop temperature activity factor.\nUse the verified formula:\n\\[F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\\]\nIn calculation-friendly notation, this is:\nF_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M).\nSubstitute the relevant values:\nTd=33.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless.\nThe computed value is 0.802 fraction.\nTherefore, the final numeric answer is \\boxed{0.802}.\n#### 0.802", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "wrong_formula_id": "CMF008"} {"answer": "0.265", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF036", "CMF050"], "effective_parameters": {"BFF": 1.17, "DTE": 0.265, "IE": 0.85, "PDT": 18.9, "RPE": 0.6}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_branch_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "parameters": {"BFF": 1.17, "DTE": 0.265, "IE": 0.85, "PDT": 18.9, "RPE": 0.6}, "problem": "For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless.\nThe computed value is 0.265 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.265}.\n#### 0.265", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF036"} {"answer": "0.687", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF023", "CMF020"], "effective_parameters": {"WaterC": 0.617}, "family": "methane_emission", "formula": {"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)", "id": "CMF025", "latex": "WI=0.49e^{3.88W_c-5.4W_c^2}", "name": "水分分解因子", "name_en": "Water-content decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF025_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]", "knowledge_formula_expr": "Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "parameters": {"WaterC": 0.617}, "problem": "Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Water-content decomposition factor.\nUse the verified formula:\n\\[WI=0.49e^{3.88W_c-5.4W_c^2}\\]\nIn calculation-friendly notation, this is:\nWI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).\nSubstitute the relevant values:\nWaterC=0.617 fraction.\nThe computed value is 0.687 fraction.\nTherefore, the final numeric answer is \\boxed{0.687}.\n#### 0.687", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF023"} {"answer": "0.329", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF025", "CMF023"], "effective_parameters": {"CH4RiceEfC": 0.39, "W": 5043.0, "Wmax": 10266.0}, "family": "methane_emission", "formula": {"expression": "CH4RiceEfC * (1 - W / Wmax) ** 0.25", "id": "CMF019", "latex": "F_p=C(1-W/W_{max})^{0.25}", "name": "植株传输因子", "name_en": "Plant methane transport factor", "precision": 3, "unit": "fraction"}, "id": "CMF019_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]", "knowledge_formula_expr": "Formula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Plant methane transport factor (植株传输因子).\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Power attenuation", "mode": "direct", "parameters": {"CH4RiceEfC": 0.39, "W": 5043.0, "Wmax": 10266.0}, "problem": "A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Plant methane transport factor (植株传输因子).\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Plant methane transport factor.\nUse the verified formula:\n\\[F_p=C(1-W/W_{max})^{0.25}\\]\nIn calculation-friendly notation, this is:\nFp = CH4RiceEfC * (1 - W / Wmax) ** 0.25.\nSubstitute the relevant values:\nCH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha.\nThe computed value is 0.329 fraction.\nTherefore, the final numeric answer is \\boxed{0.329}.\n#### 0.329", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF025"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF023", "CMF016"], "effective_parameters": {"Eh": -204.5}, "family": "methane_emission", "formula": {"expression": "if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))", "id": "CMF018", "latex": "F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}", "name": "氧化还原电位因子", "name_en": "Redox potential factor", "precision": 3, "unit": "fraction"}, "id": "CMF018_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]", "knowledge_formula_expr": "Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise exponential", "mode": "direct", "parameters": {"Eh": -204.5}, "problem": "Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Redox potential factor.\nUse the verified formula:\n\\[F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\\]\nIn calculation-friendly notation, this is:\nFEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150)).\nSubstitute the relevant values:\nEh=-204.5 mV.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF023"} {"answer": "0.637", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF058", "CMF060"], "effective_parameters": {"I50A": 483.6, "I50B": 435.8, "TT": 912.5, "Tsum": 2171.0, "fSolarMax": 0.95, "fSolar_water": 0.68}, "family": "environmental_response", "formula": {"expression": "min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)", "id": "CMF059", "latex": "fSolar=\\min(f_1,f_2)F_{solar,water}", "name": "环境响应光截获比例", "name_en": "environmental-response solar interception fraction", "precision": 3, "unit": "fraction"}, "id": "CMF059_chain_2_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]", "knowledge_formula_expr": "Formula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response solar interception fraction (环境响应光截获比例).\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Double logistic", "mode": "chain_2", "parameters": {"I50A": 483.6, "I50B": 435.8, "TT": 912.5, "Tsum": 2171.0, "fSolarMax": 0.95, "fSolar_water": 0.68}, "problem": "Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response solar interception fraction (环境响应光截获比例).\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response solar interception fraction.\nUse the verified formula:\n\\[fSolar=\\min(f_1,f_2)F_{solar,water}\\]\nIn calculation-friendly notation, this is:\nfSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1).\nSubstitute the relevant values:\nfSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction.\nThe computed value is 0.637 fraction.\nTherefore, the final numeric answer is \\boxed{0.637}.\n#### 0.637", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF058"} {"answer": "0.007", "benchmark_version": "cropmath-v1", "change": {"new_value": 36.0, "old_value": 32.3, "parameter": "t_soil"}, "distractor_formula_ids": ["CMF025", "CMF022"], "effective_parameters": {"P": 2.174, "Wr": 735.6, "t_soil": 36.0}, "family": "methane_emission", "formula": {"expression": "min(0.7 * ln(t_soil) / Wr, 0.9) * P", "id": "CMF016", "latex": "E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P", "name": "甲烷冒泡排放", "name_en": "Methane ebullition emission", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF016_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]", "knowledge_formula_expr": "Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Logarithmic minimum", "mode": "sensitivity", "parameters": {"P": 2.174, "Wr": 735.6, "t_soil": 32.3}, "problem": "In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane ebullition emission.\nUse the verified formula:\n\\[E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\\]\nIn calculation-friendly notation, this is:\nEbl = min(0.7 * ln(t_soil) / Wr, 0.9) * P.\nSubstitute the relevant values:\nP=2.174 g CH4/m2/day, t_soil=36.0 C, Wr=735.6 kg/ha.\nThe computed value is 0.007 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.007}.\n#### 0.007", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF025"} {"answer": "1.898", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF001", "CMF008"], "effective_parameters": {"Day_Length_hour": 15.26, "F_CO2": 1.25, "F_T": 0.73, "F_Water": 0.49, "Photo_Rate_opt": 6.44}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]", "knowledge_formula_expr": "Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "parameters": {"Day_Length_hour": 15.26, "F_CO2": 1.25, "F_T": 0.73, "F_Water": 0.49, "Photo_Rate_opt": 6.44}, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h.\nThe computed value is 1.898 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{1.898}.\n#### 1.898", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF001"} {"answer": "195.60", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF037", "CMF047"], "effective_parameters": {"DTGA": 489.0, "Rg": 0.4}, "family": "growth_yield", "formula": {"expression": "Rg * DTGA", "id": "CMF046", "latex": "RG=R_g\\cdot DTGA", "name": "生长呼吸", "name_en": "Growth respiration", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF046_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]", "knowledge_formula_expr": "Formula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"DTGA": 489.0, "Rg": 0.4}, "problem": "In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Growth respiration.\nUse the verified formula:\n\\[RG=R_g\\cdot DTGA\\]\nIn calculation-friendly notation, this is:\nRG = Rg * DTGA.\nSubstitute the relevant values:\nRg=0.4 fraction, DTGA=489.0 kg/ha/day.\nThe computed value is 195.60 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{195.60}.\n#### 195.60", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF037"} {"answer": "20.022", "benchmark_version": "cropmath-v1", "change": {"new_value": 3240.2, "old_value": 3600.2, "parameter": "W"}, "distractor_formula_ids": ["CMF022", "CMF018"], "effective_parameters": {"SI": 0.65, "VI": 0.7, "W": 3240.2}, "family": "methane_emission", "formula": {"expression": "0.0018 * VI * SI * W ** 1.25", "id": "CMF015", "latex": "C_r=0.0018VI\\cdot SI\\cdot W^{1.25}", "name": "根系分泌碳底物", "name_en": "Root-exudate carbon substrate", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF015_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]", "knowledge_formula_expr": "Formula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Root-exudate carbon substrate (根系分泌碳底物).\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Power product", "mode": "sensitivity", "parameters": {"SI": 0.65, "VI": 0.7, "W": 3600.2}, "problem": "In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root-exudate carbon substrate (根系分泌碳底物).\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root-exudate carbon substrate.\nUse the verified formula:\n\\[C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\\]\nIn calculation-friendly notation, this is:\nCr = 0.0018 * VI * SI * W ** 1.25.\nSubstitute the relevant values:\nVI=0.7 fraction, SI=0.65 fraction, W=3240.2 kg/ha.\nThe computed value is 20.022 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{20.022}.\n#### 20.022", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF022"} {"answer": "0.580", "benchmark_version": "cropmath-v1", "change": {"new_value": 13.5, "old_value": 11.7, "parameter": "Tmn"}, "distractor_formula_ids": ["CMF012", "CMF006"], "effective_parameters": {"Biom_plant_yesterday": 279.1, "Coef_Rm": 0.002, "DVI": 0.51, "Tmn": 13.5}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "parameters": {"Biom_plant_yesterday": 279.1, "Coef_Rm": 0.002, "DVI": 0.51, "Tmn": 11.7}, "problem": "Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=13.5 C, DVI=0.51 dimensionless.\nThe computed value is 0.580 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.580}.\n#### 0.580", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "wrong_formula_id": "CMF012"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF028", "CMF038"], "effective_parameters": {"T": 30.6, "Tb": 10.0, "Tmax": 45.0, "Tol": 24.0, "Tou": 34.0}, "family": "growth_yield", "formula": {"expression": "if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Temperature response factor.\nUse the verified formula:\n\\[F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\\]\nIn calculation-friendly notation, this is:\nF_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative photoperiod effectiveness (相对光周期效应).\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Relative photoperiod effectiveness.\nUse the verified formula:\n\\[RPE=1,\\;1-PS(P-P_o)^2,\\;0\\]\nIn calculation-friendly notation, this is:\nRPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0.\nSubstitute the relevant values:\nP=10.17 h, Po=10.5 h, PS=0.121 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF048"} {"answer": "0.196", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF005", "CMF007"], "effective_parameters": {"Coef_Plant_TN_uptake": 0.058, "DVI": 0.29, "NPP": 13.98, "N_requirement_max_day": 1.44, "N_shoot": 4.79, "Ratio_N_C_photo_product": 0.014}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)", "id": "CMF004", "latex": "N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)", "name": "作物需氮量", "name_en": "carbon-nitrogen cycling crop nitrogen demand", "precision": 3, "unit": "g N/m2/day"}, "id": "CMF004_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]", "knowledge_formula_expr": "Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "parameters": {"Coef_Plant_TN_uptake": 0.058, "DVI": 0.29, "NPP": 13.98, "N_requirement_max_day": 1.44, "N_shoot": 4.79, "Ratio_N_C_photo_product": 0.014}, "problem": "A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop nitrogen demand.\nUse the verified formula:\n\\[N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\\]\nIn calculation-friendly notation, this is:\nN_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP).\nSubstitute the relevant values:\nN_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless.\nThe computed value is 0.196 g N/m2/day.\nTherefore, the final numeric answer is \\boxed{0.196}.\n#### 0.196", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "wrong_formula_id": "CMF005"} {"answer": "1.028", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF011", "CMF003"], "effective_parameters": {"Coef_T_M": 0.52, "Td": 25.7, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)", "id": "CMF013", "latex": "F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}", "name": "作物温度活性因子", "name_en": "carbon-nitrogen cycling crop temperature activity factor", "precision": 3, "unit": "response factor"}, "id": "CMF013_branch_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]", "knowledge_formula_expr": "Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Beta temperature response", "mode": "branch", "parameters": {"Coef_T_M": 0.52, "Td": 25.7, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "problem": "For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop temperature activity factor.\nUse the verified formula:\n\\[F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\\]\nIn calculation-friendly notation, this is:\nF_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M).\nSubstitute the relevant values:\nTd=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless.\nThe computed value is 1.028 fraction.\nTherefore, the final numeric answer is \\boxed{1.028}.\n#### 1.028", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 1, "wrong_formula_id": "CMF011"} {"answer": "0.278", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF055", "CMF058"], "effective_parameters": {"Tbase": 6.6, "Tmean": 12.5, "Topt": 27.8}, "family": "environmental_response", "formula": {"expression": "if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)", "id": "CMF060", "latex": "F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)", "name": "环境响应温度响应", "name_en": "environmental-response temperature response", "precision": 3, "unit": "fraction"}, "id": "CMF060_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]", "knowledge_formula_expr": "Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise linear", "mode": "direct", "parameters": {"Tbase": 6.6, "Tmean": 12.5, "Topt": 27.8}, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response temperature response.\nUse the verified formula:\n\\[F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\\]\nIn calculation-friendly notation, this is:\nF_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0).\nSubstitute the relevant values:\nTmean=12.5 C, Tbase=6.6 C, Topt=27.8 C.\nThe computed value is 0.278 fraction.\nTherefore, the final numeric answer is \\boxed{0.278}.\n#### 0.278", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF055"} {"answer": "0.0020", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.45, "old_value": 0.5, "parameter": "NNI"}, "distractor_formula_ids": ["CMF037", "CMF041"], "effective_parameters": {"NNI": 0.45, "RP": 0.0044, "WDF": 0.92}, "family": "growth_yield", "formula": {"expression": "RP * min(NNI, WDF)", "id": "CMF052", "latex": "R=RP\\cdot\\min(NNI,WDF)", "name": "实际叶面积相对增长率", "name_en": "Actual LAI relative growth rate", "precision": 4, "unit": "d-1"}, "id": "CMF052_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_name": "Relevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "parameters": {"NNI": 0.5, "RP": 0.0044, "WDF": 0.92}, "problem": "In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual LAI relative growth rate.\nUse the verified formula:\n\\[R=RP\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nR = RP * min(NNI, WDF).\nSubstitute the relevant values:\nRP=0.0044 d-1, NNI=0.45 fraction, WDF=0.92 fraction.\nThe computed value is 0.0020 d-1.\nTherefore, the final numeric answer is \\boxed{0.0020}.\n#### 0.0020", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF037"} {"answer": "0.867", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF004", "CMF005"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.574, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop water stress factor.\nUse the verified formula:\n\\[F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\\]\nIn calculation-friendly notation, this is:\nF_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子).\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil pH decomposition factor.\nUse the verified formula:\n\\[F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\\]\nIn calculation-friendly notation, this is:\nF_pH = 1 / (1 + exp(2.5 * (5 - pH))).\nSubstitute the relevant values:\npH=7.85 pH.\nThe computed value is 0.999 fraction.\nTherefore, the final numeric answer is \\boxed{0.999}.\n#### 0.999", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "wrong_formula_id": "CMF007"} {"answer": "83.94", "benchmark_version": "cropmath-v1", "change": {"new_value": 299.8, "old_value": 260.7, "parameter": "DTGA"}, "distractor_formula_ids": ["CMF035", "CMF049"], "effective_parameters": {"DTGA": 299.8, "Rg": 0.28}, "family": "growth_yield", "formula": {"expression": "Rg * DTGA", "id": "CMF046", "latex": "RG=R_g\\cdot DTGA", "name": "生长呼吸", "name_en": "Growth respiration", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF046_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]", "knowledge_formula_expr": "Formula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product", "mode": "sensitivity", "parameters": {"DTGA": 260.7, "Rg": 0.28}, "problem": "Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Growth respiration.\nUse the verified formula:\n\\[RG=R_g\\cdot DTGA\\]\nIn calculation-friendly notation, this is:\nRG = Rg * DTGA.\nSubstitute the relevant values:\nRg=0.28 fraction, DTGA=299.8 kg/ha/day.\nThe computed value is 83.94 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{83.94}.\n#### 83.94", "split": "dev", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF035"} {"answer": "0.960", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF047", "CMF027"], "effective_parameters": {"T": 26.9, "TS": 1.35, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "family": "growth_yield", "formula": {"expression": "(((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS", "id": "CMF051", "latex": "RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}", "name": "相对热效应", "name_en": "Relative thermal effectiveness", "precision": 3, "unit": "fraction"}, "id": "CMF051_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]", "knowledge_formula_expr": "Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Beta temperature response", "mode": "direct", "parameters": {"T": 26.9, "TS": 1.35, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "problem": "Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Relative thermal effectiveness.\nUse the verified formula:\n\\[RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\\]\nIn calculation-friendly notation, this is:\nRTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS.\nSubstitute the relevant values:\nT=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless.\nThe computed value is 0.960 fraction.\nTherefore, the final numeric answer is \\boxed{0.960}.\n#### 0.960", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF047"} {"answer": "8.259", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF015", "CMF024"], "effective_parameters": {"Com": 7.123, "Cr": 61.55, "FEh": 0.478, "TI": 0.924}, "family": "methane_emission", "formula": {"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))", "id": "CMF022", "latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))", "name": "甲烷产生量", "name_en": "Methane production", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF022_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]", "knowledge_formula_expr": "Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "parameters": {"Com": 7.123, "Cr": 61.55, "FEh": 0.478, "TI": 0.924}, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane production.\nUse the verified formula:\n\\[P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\\]\nIn calculation-friendly notation, this is:\nP = max(0, 0.27 * FEh * (TI * Cr + Com)).\nSubstitute the relevant values:\nFEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day.\nThe computed value is 8.259 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{8.259}.\n#### 8.259", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF015"} {"answer": "0.326", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF015", "CMF024"], "effective_parameters": {"Q10": 3.52, "t_soil": 21.1}, "family": "methane_emission", "formula": {"expression": "Q10^((min(t_soil,30)-30)/10)", "id": "CMF023", "latex": "TI=Q_{10}^{(t_{sl}-30)/10}", "name": "甲烷温度指数", "name_en": "Methane temperature index", "precision": 3, "unit": "multiplier"}, "id": "CMF023_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]", "knowledge_formula_expr": "Formula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "parameters": {"Q10": 3.52, "t_soil": 21.1}, "problem": "In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane temperature index.\nUse the verified formula:\n\\[TI=Q_{10}^{(t_{sl}-30)/10}\\]\nIn calculation-friendly notation, this is:\nTI = Q10^((min(t_soil,30)-30)/10).\nSubstitute the relevant values:\nQ10=3.52 dimensionless, t_soil=21.1 C.\nThe computed value is 0.326 fraction.\nTherefore, the final numeric answer is \\boxed{0.326}.\n#### 0.326", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF015"} {"answer": "0.11", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF031", "CMF033"], "effective_parameters": {"B_above": 17858.0, "PDT": 34.4, "Y_grain": 7851.0}, "family": "growth_yield", "formula": {"expression": "(Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))", "id": "CMF042", "latex": "PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}", "name": "穗分配指数", "name_en": "Panicle partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF042_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]", "knowledge_formula_expr": "Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "parameters": {"B_above": 17858.0, "PDT": 34.4, "Y_grain": 7851.0}, "problem": "Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Panicle partitioning index.\nUse the verified formula:\n\\[PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\\]\nIn calculation-friendly notation, this is:\nPIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39))).\nSubstitute the relevant values:\nY_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days.\nThe computed value is 0.11 fraction.\nTherefore, the final numeric answer is \\boxed{0.11}.\n#### 0.11", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF031"} {"answer": "0.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF061", "CMF057"], "effective_parameters": {"Tbase": 6.8, "Tmean": 3.0}, "family": "environmental_response", "formula": {"expression": "max(Tmean - Tbase, 0)", "id": "CMF056", "latex": "dTT=\\max(T_{mean}-T_{base},0)", "name": "每日有效积温", "name_en": "environmental-response daily thermal time", "precision": 2, "unit": "C day"}, "id": "CMF056_branch_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]", "knowledge_formula_expr": "Formula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Maximum linear", "mode": "branch", "parameters": {"Tbase": 6.8, "Tmean": 3.0}, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily thermal time.\nUse the verified formula:\n\\[dTT=\\max(T_{mean}-T_{base},0)\\]\nIn calculation-friendly notation, this is:\ndTT = max(Tmean - Tbase, 0).\nSubstitute the relevant values:\nTmean=3.0 C, Tbase=6.8 C.\nThe computed value is 0.00 C day.\nTherefore, the final numeric answer is \\boxed{0.00}.\n#### 0.00", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF061"} {"answer": "29.939", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF002", "CMF010"], "effective_parameters": {"Coef_alfa": 27.9, "Coef_beta": 146.2, "Leaf_N": 1.58, "PAR": 309.5}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "(Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)", "id": "CMF007", "latex": "P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}", "name": "最适光合速率", "name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "precision": 3, "unit": "umol CO2/m2/s"}, "id": "CMF007_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]", "knowledge_formula_expr": "Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Michaelis-Menten response", "mode": "chain_2", "parameters": {"Coef_alfa": 27.9, "Coef_beta": 146.2, "Leaf_N": 1.58, "PAR": 309.5}, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling optimal photosynthetic rate.\nUse the verified formula:\n\\[P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\\]\nIn calculation-friendly notation, this is:\nPhoto_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta).\nSubstitute the relevant values:\nCoef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2.\nThe computed value is 29.939 umol CO2/m2/s.\nTherefore, the final numeric answer is \\boxed{29.939}.\n#### 29.939", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF002"} {"answer": "0.771", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF031", "CMF039"], "effective_parameters": {"BFF": 1.19, "DTE": 0.771, "IE": 1.24, "PDT": 22.1, "RPE": 0.81}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "parameters": {"BFF": 1.19, "DTE": 0.771, "IE": 1.24, "PDT": 22.1, "RPE": 0.81}, "problem": "In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless.\nThe computed value is 0.771 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.771}.\n#### 0.771", "split": "dev", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF031"} {"answer": "5.724", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.004, "old_value": 0.0037, "parameter": "Coef_Rm"}, "distractor_formula_ids": ["CMF013", "CMF002"], "effective_parameters": {"Biom_plant_yesterday": 941.4, "Coef_Rm": 0.004, "DVI": 0.78, "Tmn": 19.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "parameters": {"Biom_plant_yesterday": 941.4, "Coef_Rm": 0.0037, "DVI": 0.78, "Tmn": 19.0}, "problem": "Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.004 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless.\nThe computed value is 5.724 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{5.724}.\n#### 5.724", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "wrong_formula_id": "CMF013"} {"answer": "0.842", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF010", "CMF004"], "effective_parameters": {"BD": 1.26, "Saturate_Water": 0.48, "Soil_Water": 0.239}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)", "id": "CMF012", "latex": "F_W=0.49e^{3.88W-5.4W^2}", "name": "土壤碳分解水分因子", "name_en": "carbon-nitrogen cycling soil carbon water factor", "precision": 3, "unit": "fraction"}, "id": "CMF012_branch_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]", "knowledge_formula_expr": "Formula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子).\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "parameters": {"BD": 1.26, "Saturate_Water": 0.48, "Soil_Water": 0.239}, "problem": "Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子).\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil carbon water factor.\nUse the verified formula:\n\\[F_W=0.49e^{3.88W-5.4W^2}\\]\nIn calculation-friendly notation, this is:\nF_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2).\nSubstitute the relevant values:\nSoil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3.\nThe computed value is 0.842 fraction.\nTherefore, the final numeric answer is \\boxed{0.842}.\n#### 0.842", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "wrong_formula_id": "CMF010"} {"answer": "0.978", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF021", "CMF016"], "effective_parameters": {"WaterC": 0.393}, "family": "methane_emission", "formula": {"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)", "id": "CMF025", "latex": "WI=0.49e^{3.88W_c-5.4W_c^2}", "name": "水分分解因子", "name_en": "Water-content decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF025_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]", "knowledge_formula_expr": "Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "parameters": {"WaterC": 0.393}, "problem": "A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Water-content decomposition factor.\nUse the verified formula:\n\\[WI=0.49e^{3.88W_c-5.4W_c^2}\\]\nIn calculation-friendly notation, this is:\nWI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).\nSubstitute the relevant values:\nWaterC=0.393 fraction.\nThe computed value is 0.978 fraction.\nTherefore, the final numeric answer is \\boxed{0.978}.\n#### 0.978", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF021"} {"answer": "1.177", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF005"], "effective_parameters": {"CO2": 459.0, "Coef_B_CO2": 0.59}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "1 + Coef_B_CO2 * ln(CO2 / 340)", "id": "CMF001", "latex": "F_{CO2}=1+B\\ln(CO2/340)", "name": "作物二氧化碳施肥因子", "name_en": "carbon-nitrogen cycling CO2 fertilization factor", "precision": 3, "unit": "multiplier"}, "id": "CMF001_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]", "knowledge_formula_expr": "Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Logarithmic", "mode": "direct", "parameters": {"CO2": 459.0, "Coef_B_CO2": 0.59}, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling CO2 fertilization factor.\nUse the verified formula:\n\\[F_{CO2}=1+B\\ln(CO2/340)\\]\nIn calculation-friendly notation, this is:\nF_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340).\nSubstitute the relevant values:\nCO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless.\nThe computed value is 1.177 fraction.\nTherefore, the final numeric answer is \\boxed{1.177}.\n#### 1.177", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF009"} {"answer": "38.77", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF051", "CMF047"], "effective_parameters": {"Am": 41.5, "FCO2": 1.05, "FPA": 0.93, "FT": 0.92, "NNI": 1.04, "WDF": 1.08}, "family": "growth_yield", "formula": {"expression": "Am * FCO2 * FPA * FT * min(NNI, WDF)", "id": "CMF026", "latex": "AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)", "name": "实际最大同化速率", "name_en": "Actual maximum assimilation rate", "precision": 2, "unit": "kg CO2/ha/h"}, "id": "CMF026_chain_2_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_name": "Relevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "parameters": {"Am": 41.5, "FCO2": 1.05, "FPA": 0.93, "FT": 0.92, "NNI": 1.04, "WDF": 1.08}, "problem": "A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual maximum assimilation rate.\nUse the verified formula:\n\\[AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nAMAX = Am * FCO2 * FPA * FT * min(NNI, WDF).\nSubstitute the relevant values:\nAm=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction.\nThe computed value is 38.77 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{38.77}.\n#### 38.77", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF051"} {"answer": "5978.8", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF058", "CMF055"], "effective_parameters": {"Biomass": 16159.0, "HI": 0.37}, "family": "environmental_response", "formula": {"expression": "Biomass * HI", "id": "CMF062", "latex": "Yield=Biomass\\cdot HI", "name": "环境响应产量", "name_en": "environmental-response yield", "precision": 1, "unit": "kg/ha"}, "id": "CMF062_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]", "knowledge_formula_expr": "Formula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha", "knowledge_name": "Relevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"Biomass": 16159.0, "HI": 0.37}, "problem": "A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response yield.\nUse the verified formula:\n\\[Yield=Biomass\\cdot HI\\]\nIn calculation-friendly notation, this is:\nYield = Biomass * HI.\nSubstitute the relevant values:\nBiomass=16159.0 kg/ha, HI=0.37 fraction.\nThe computed value is 5978.8 kg/ha.\nTherefore, the final numeric answer is \\boxed{5978.8}.\n#### 5978.8", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF058"} {"answer": "8.285", "benchmark_version": "cropmath-v1", "change": {"new_value": 753.4, "old_value": 684.9, "parameter": "Biom_shoot"}, "distractor_formula_ids": ["CMF006", "CMF011"], "effective_parameters": {"Biom_root": 24.1, "Biom_shoot": 753.4, "Plant_TN_uptake": 8.55}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)", "id": "CMF005", "latex": "N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}", "name": "地上部氮分配量", "name_en": "carbon-nitrogen cycling shoot nitrogen allocation", "precision": 3, "unit": "g N/m2"}, "id": "CMF005_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]", "knowledge_formula_expr": "Formula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量).\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "parameters": {"Biom_root": 24.1, "Biom_shoot": 684.9, "Plant_TN_uptake": 8.55}, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量).\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling shoot nitrogen allocation.\nUse the verified formula:\n\\[N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\\]\nIn calculation-friendly notation, this is:\nN_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root).\nSubstitute the relevant values:\nPlant_TN_uptake=8.55 g N/m2, Biom_shoot=753.4 g C/m2, Biom_root=24.1 g C/m2.\nThe computed value is 8.285 g N/m2.\nTherefore, the final numeric answer is \\boxed{8.285}.\n#### 8.285", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "wrong_formula_id": "CMF006"} {"answer": "439.26", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF047", "CMF049"], "effective_parameters": {"DTGA": 565.5, "RG": 53.7, "RM": 94.5, "b": 0.05}, "family": "growth_yield", "formula": {"expression": "(DTGA - RM - RG) / (1 - b)", "id": "CMF033", "latex": "GCR=\\frac{DTGA-RM-RG}{1-b}", "name": "日干物质增量", "name_en": "Daily dry matter increment", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF033_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]", "knowledge_formula_expr": "Formula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "parameters": {"DTGA": 565.5, "RG": 53.7, "RM": 94.5, "b": 0.05}, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily dry matter increment.\nUse the verified formula:\n\\[GCR=\\frac{DTGA-RM-RG}{1-b}\\]\nIn calculation-friendly notation, this is:\nGCR = (DTGA - RM - RG) / (1 - b).\nSubstitute the relevant values:\nDTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction.\nThe computed value is 439.26 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{439.26}.\n#### 439.26", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF047"} {"answer": "12.90", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF058", "CMF059"], "effective_parameters": {"Tbase": 10.6, "Tmean": 23.5}, "family": "environmental_response", "formula": {"expression": "max(Tmean - Tbase, 0)", "id": "CMF056", "latex": "dTT=\\max(T_{mean}-T_{base},0)", "name": "每日有效积温", "name_en": "environmental-response daily thermal time", "precision": 2, "unit": "C day"}, "id": "CMF056_branch_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]", "knowledge_formula_expr": "Formula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Maximum linear", "mode": "branch", "parameters": {"Tbase": 10.6, "Tmean": 23.5}, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily thermal time.\nUse the verified formula:\n\\[dTT=\\max(T_{mean}-T_{base},0)\\]\nIn calculation-friendly notation, this is:\ndTT = max(Tmean - Tbase, 0).\nSubstitute the relevant values:\nTmean=23.5 C, Tbase=10.6 C.\nThe computed value is 12.90 C day.\nTherefore, the final numeric answer is \\boxed{12.90}.\n#### 12.90", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF058"} {"answer": "0.331", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF018", "CMF017"], "effective_parameters": {"OMS": 1246.5, "SI": 1.15, "TI": 0.28, "WI": 0.275}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.003 * OMS", "id": "CMF021", "latex": "OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS", "name": "土壤有机质碳分解量", "name_en": "Soil organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF021_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]", "knowledge_formula_expr": "Formula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "parameters": {"OMS": 1246.5, "SI": 1.15, "TI": 0.28, "WI": 0.275}, "problem": "In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil organic matter carbon decomposition.\nUse the verified formula:\n\\[OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\\]\nIn calculation-friendly notation, this is:\nOMSC = WI * SI * TI * 0.003 * OMS.\nSubstitute the relevant values:\nWI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2.\nThe computed value is 0.331 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.331}.\n#### 0.331", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF018"} {"answer": "6.319", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF018", "CMF016"], "effective_parameters": {"CH4RiceEfC": 0.208, "Ebl": 4.391, "P": 11.616, "W": 5453.7, "Wmax": 9178.0}, "family": "methane_emission", "formula": {"expression": "Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25", "id": "CMF017", "latex": "E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}", "name": "甲烷总排放", "name_en": "Total methane emission", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF017_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]", "knowledge_formula_expr": "Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Chained sum", "mode": "direct", "parameters": {"CH4RiceEfC": 0.208, "Ebl": 4.391, "P": 11.616, "W": 5453.7, "Wmax": 9178.0}, "problem": "In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Total methane emission.\nUse the verified formula:\n\\[E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\\]\nIn calculation-friendly notation, this is:\nE = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25.\nSubstitute the relevant values:\nEbl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha.\nThe computed value is 6.319 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{6.319}.\n#### 6.319", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF018"} {"answer": "0.345", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF054", "CMF049"], "effective_parameters": {"BFF": 0.67, "DTE": 0.515, "IE": 1.24, "PDT": 34.5, "RPE": 0.49}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_branch_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "parameters": {"BFF": 0.67, "DTE": 0.515, "IE": 1.24, "PDT": 34.5, "RPE": 0.49}, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless.\nThe computed value is 0.345 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.345}.\n#### 0.345", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF054"} {"answer": "27.20", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF019", "CMF016"], "effective_parameters": {"Tair": 30.0}, "family": "methane_emission", "formula": {"expression": "4.4 + 0.76 * Tair", "id": "CMF024", "latex": "T_{soil}=4.4+0.76T_{air}", "name": "土壤温度估算", "name_en": "Soil temperature from air temperature", "precision": 2, "unit": "C"}, "id": "CMF024_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]", "knowledge_formula_expr": "Formula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_name": "Relevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear", "mode": "direct", "parameters": {"Tair": 30.0}, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil temperature from air temperature.\nUse the verified formula:\n\\[T_{soil}=4.4+0.76T_{air}\\]\nIn calculation-friendly notation, this is:\nTsoil = 4.4 + 0.76 * Tair.\nSubstitute the relevant values:\nTair=30.0 C.\nThe computed value is 27.20 C.\nTherefore, the final numeric answer is \\boxed{27.20}.\n#### 27.20", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF019"} {"answer": "0.700", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF004", "CMF002"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.178, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop water stress factor.\nUse the verified formula:\n\\[F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\\]\nIn calculation-friendly notation, this is:\nF_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]", "knowledge_formula_expr": "Formula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product", "mode": "sensitivity", "parameters": {"DTGA": 369.5, "Rg": 0.36}, "problem": "Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Growth respiration.\nUse the verified formula:\n\\[RG=R_g\\cdot DTGA\\]\nIn calculation-friendly notation, this is:\nRG = Rg * DTGA.\nSubstitute the relevant values:\nRg=0.41 fraction, DTGA=369.5 kg/ha/day.\nThe computed value is 151.50 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{151.50}.\n#### 151.50", "split": "dev", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF033"} {"answer": "0.475", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.22, "old_value": 1.35, "parameter": "ARID"}, "distractor_formula_ids": ["CMF062", "CMF056"], "effective_parameters": {"ARID": 1.22, "S_Water": 0.43}, "family": "environmental_response", "formula": {"expression": "max(0, 1 - S_Water * ARID)", "id": "CMF061", "latex": "F_W=\\max(0,1-S_WARID)", "name": "环境响应水分响应", "name_en": "environmental-response water stress response", "precision": 3, "unit": "response factor"}, "id": "CMF061_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]", "knowledge_formula_expr": "Formula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response water stress response (环境响应水分响应).\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "parameters": {"ARID": 1.35, "S_Water": 0.43}, "problem": "Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response water stress response (环境响应水分响应).\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response water stress response.\nUse the verified formula:\n\\[F_W=\\max(0,1-S_WARID)\\]\nIn calculation-friendly notation, this is:\nF_Water = max(0, 1 - S_Water * ARID).\nSubstitute the relevant values:\nARID=1.22 dimensionless, S_Water=0.43 dimensionless.\nThe computed value is 0.475 fraction.\nTherefore, the final numeric answer is \\boxed{0.475}.\n#### 0.475", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF062"} {"answer": "0.08", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF051", "CMF037"], "effective_parameters": {"PDT": 51.0}, "family": "growth_yield", "formula": {"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)", "id": "CMF043", "latex": "PIRO=1-PISH", "name": "地下部分配指数", "name_en": "Root partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF043_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "parameters": {"PDT": 51.0}, "problem": "Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root partitioning index.\nUse the verified formula:\n\\[PIRO=1-PISH\\]\nIn calculation-friendly notation, this is:\nPIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).\nSubstitute the relevant values:\nPDT=51.0 physiological days.\nThe computed value is 0.08 fraction.\nTherefore, the final numeric answer is \\boxed{0.08}.\n#### 0.08", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF051"} {"answer": "0.0077", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF042", "CMF041"], "effective_parameters": {"PDT": 14.5}, "family": "growth_yield", "formula": {"expression": "0.0091 - 0.0001 * PDT", "id": "CMF049", "latex": "RM(T_o)=0.0091-0.0001PDT", "name": "最适温度维持呼吸系数", "name_en": "Maintenance respiration coefficient at optimum temperature", "precision": 4, "unit": "d-1"}, "id": "CMF049_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_name": "Relevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear decline", "mode": "direct", "parameters": {"PDT": 14.5}, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration coefficient at optimum temperature.\nUse the verified formula:\n\\[RM(T_o)=0.0091-0.0001PDT\\]\nIn calculation-friendly notation, this is:\nRMTo = 0.0091 - 0.0001 * PDT.\nSubstitute the relevant values:\nPDT=14.5 physiological days.\nThe computed value is 0.0077 d-1.\nTherefore, the final numeric answer is \\boxed{0.0077}.\n#### 0.0077", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF042"} {"answer": "0.34", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF052", "CMF031"], "effective_parameters": {"PDT": 50.0, "a": 0.049}, "family": "growth_yield", "formula": {"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))", "id": "CMF030", "latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)", "name": "生理年龄因子", "name_en": "Physiological age factor", "precision": 2, "unit": "fraction"}, "id": "CMF030_branch_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]", "knowledge_formula_expr": "Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise exponential decay", "mode": "branch", "parameters": {"PDT": 50.0, "a": 0.049}, "problem": "Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Physiological age factor.\nUse the verified formula:\n\\[F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\\]\nIn calculation-friendly notation, this is:\nFPA = if PDT<28: 1; else: exp(-a*(PDT-28)).\nSubstitute the relevant values:\nPDT=50.0 physiological days, a=0.049 dimensionless.\nThe computed value is 0.34 fraction.\nTherefore, the final numeric answer is \\boxed{0.34}.\n#### 0.34", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF052"} {"answer": "14.04", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.08, "old_value": 0.083, "parameter": "PS"}, "distractor_formula_ids": ["CMF029", "CMF045"], "effective_parameters": {"PS": 0.08, "Po": 10.5}, "family": "growth_yield", "formula": {"expression": "Po + sqrt(1 / PS)", "id": "CMF040", "latex": "P_c=P_o+\\sqrt{1/PS}", "name": "临界日长", "name_en": "Critical daylength", "precision": 2, "unit": "h"}, "id": "CMF040_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]", "knowledge_formula_expr": "Formula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_name": "Relevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Power response", "mode": "sensitivity", "parameters": {"PS": 0.083, "Po": 10.5}, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Critical daylength.\nUse the verified formula:\n\\[P_c=P_o+\\sqrt{1/PS}\\]\nIn calculation-friendly notation, this is:\nPc = Po + sqrt(1 / PS).\nSubstitute the relevant values:\nPo=10.5 h, PS=0.08 dimensionless.\nThe computed value is 14.04 h.\nTherefore, the final numeric answer is \\boxed{14.04}.\n#### 14.04", "split": "dev", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF029"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF025", "CMF018"], "effective_parameters": {"Q10": 2.2, "t_soil": 39.9}, "family": "methane_emission", "formula": {"expression": "Q10^((min(t_soil,30)-30)/10)", "id": "CMF023", "latex": "TI=Q_{10}^{(t_{sl}-30)/10}", "name": "甲烷温度指数", "name_en": "Methane temperature index", "precision": 3, "unit": "multiplier"}, "id": "CMF023_branch_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]", "knowledge_formula_expr": "Formula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Q10 with temperature clipping", "mode": "branch", "parameters": {"Q10": 2.2, "t_soil": 39.9}, "problem": "For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane temperature index.\nUse the verified formula:\n\\[TI=Q_{10}^{(t_{sl}-30)/10}\\]\nIn calculation-friendly notation, this is:\nTI = Q10^((min(t_soil,30)-30)/10).\nSubstitute the relevant values:\nQ10=2.2 dimensionless, t_soil=39.9 C.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF025"} {"answer": "1.18", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF052", "CMF051"], "effective_parameters": {"Cx": 419.0, "beta": 0.86}, "family": "growth_yield", "formula": {"expression": "1 + beta * ln(Cx / 340)", "id": "CMF029", "latex": "F_{CO2}=1+\\beta\\ln(C_x/340)", "name": "二氧化碳影响因子", "name_en": "CO2 response factor", "precision": 2, "unit": "multiplier"}, "id": "CMF029_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]", "knowledge_formula_expr": "Formula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Logarithmic", "mode": "direct", "parameters": {"Cx": 419.0, "beta": 0.86}, "problem": "Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as CO2 response factor.\nUse the verified formula:\n\\[F_{CO2}=1+\\beta\\ln(C_x/340)\\]\nIn calculation-friendly notation, this is:\nFCO2 = 1 + beta * ln(Cx / 340).\nSubstitute the relevant values:\nCx=419.0 ppm, beta=0.86 dimensionless.\nThe computed value is 1.18 fraction.\nTherefore, the final numeric answer is \\boxed{1.18}.\n#### 1.18", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF052"} {"answer": "0.198", "benchmark_version": "cropmath-v1", "change": {"new_value": -7.0, "old_value": -6.1, "parameter": "Eh"}, "distractor_formula_ids": ["CMF023", "CMF021"], "effective_parameters": {"Eh": -7.0}, "family": "methane_emission", "formula": {"expression": "if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))", "id": "CMF018", "latex": "F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}", "name": "氧化还原电位因子", "name_en": "Redox potential factor", "precision": 3, "unit": "fraction"}, "id": "CMF018_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]", "knowledge_formula_expr": "Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "parameters": {"Eh": -6.1}, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Redox potential factor.\nUse the verified formula:\n\\[F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\\]\nIn calculation-friendly notation, this is:\nFEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150)).\nSubstitute the relevant values:\nEh=-7.0 mV.\nThe computed value is 0.198 fraction.\nTherefore, the final numeric answer is \\boxed{0.198}.\n#### 0.198", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF023"} {"answer": "71.6", "benchmark_version": "cropmath-v1", "change": {"new_value": 3.8, "old_value": 4.2, "parameter": "SDEPTH"}, "distractor_formula_ids": ["CMF033", "CMF034"], "effective_parameters": {"SDEPTH": 3.8}, "family": "growth_yield", "formula": {"expression": "45 + 7 * SDEPTH", "id": "CMF028", "latex": "EMGDD=45+7SDEPTH", "name": "出苗热量需求", "name_en": "Emergence thermal requirement", "precision": 1, "unit": "C day"}, "id": "CMF028_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]", "knowledge_formula_expr": "Formula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Linear", "mode": "sensitivity", "parameters": {"SDEPTH": 4.2}, "problem": "Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Emergence thermal requirement.\nUse the verified formula:\n\\[EMGDD=45+7SDEPTH\\]\nIn calculation-friendly notation, this is:\nEMGDD = 45 + 7 * SDEPTH.\nSubstitute the relevant values:\nSDEPTH=3.8 cm.\nThe computed value is 71.6 C day.\nTherefore, the final numeric answer is \\boxed{71.6}.\n#### 71.6", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF033"} {"answer": "10.69", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF031", "CMF027"], "effective_parameters": {"Am": 28.3, "FCO2": 1.13, "FPA": 0.91, "FT": 0.54, "NNI": 0.68, "WDF": 0.73}, "family": "growth_yield", "formula": {"expression": "Am * FCO2 * FPA * FT * min(NNI, WDF)", "id": "CMF026", "latex": "AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)", "name": "实际最大同化速率", "name_en": "Actual maximum assimilation rate", "precision": 2, "unit": "kg CO2/ha/h"}, "id": "CMF026_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_name": "Relevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "parameters": {"Am": 28.3, "FCO2": 1.13, "FPA": 0.91, "FT": 0.54, "NNI": 0.68, "WDF": 0.73}, "problem": "A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual maximum assimilation rate.\nUse the verified formula:\n\\[AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nAMAX = Am * FCO2 * FPA * FT * min(NNI, WDF).\nSubstitute the relevant values:\nAm=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction.\nThe computed value is 10.69 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{10.69}.\n#### 10.69", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF031"} {"answer": "0.962", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF045", "CMF054"], "effective_parameters": {"BFF": 1.04, "DTE": 0.925, "IE": 0.89, "PDT": 48.6, "RPE": 0.52}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_chain_2_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "parameters": {"BFF": 1.04, "DTE": 0.925, "IE": 0.89, "PDT": 48.6, "RPE": 0.52}, "problem": "A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless.\nThe computed value is 0.962 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.962}.\n#### 0.962", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF045"} {"answer": "0.07", "benchmark_version": "cropmath-v1", "change": {"new_value": 57.0, "old_value": 56.5, "parameter": "PDT"}, "distractor_formula_ids": ["CMF031", "CMF036"], "effective_parameters": {"PDT": 57.0}, "family": "growth_yield", "formula": {"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)", "id": "CMF043", "latex": "PIRO=1-PISH", "name": "地下部分配指数", "name_en": "Root partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF043_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "parameters": {"PDT": 56.5}, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root partitioning index.\nUse the verified formula:\n\\[PIRO=1-PISH\\]\nIn calculation-friendly notation, this is:\nPIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).\nSubstitute the relevant values:\nPDT=57.0 physiological days.\nThe computed value is 0.07 fraction.\nTherefore, the final numeric answer is \\boxed{0.07}.\n#### 0.07", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF031"} {"answer": "0.982", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF048", "CMF036"], "effective_parameters": {"T": 27.7, "TS": 1.07, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "family": "growth_yield", "formula": {"expression": "(((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS", "id": "CMF051", "latex": "RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}", "name": "相对热效应", "name_en": "Relative thermal effectiveness", "precision": 3, "unit": "fraction"}, "id": "CMF051_branch_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]", "knowledge_formula_expr": "Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Beta temperature response", "mode": "branch", "parameters": {"T": 27.7, "TS": 1.07, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Relative thermal effectiveness.\nUse the verified formula:\n\\[RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\\]\nIn calculation-friendly notation, this is:\nRTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS.\nSubstitute the relevant values:\nT=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless.\nThe computed value is 0.982 fraction.\nTherefore, the final numeric answer is \\boxed{0.982}.\n#### 0.982", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF048"} {"answer": "0.045", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF007", "CMF003"], "effective_parameters": {"Coef_K_Decom": 0.0057, "Coef_K_Trans": 0.086, "F_notill": 0.93, "F_pH": 0.101, "F_temp": 1.373, "F_texture": 0.899, "F_water": 0.266, "Pool": 281.7}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)", "id": "CMF009", "latex": "R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})", "name": "碳库呼吸损失", "name_en": "carbon-nitrogen cycling carbon pool respiration loss", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF009_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]", "knowledge_formula_expr": "Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Chained multiplicative decomposition", "mode": "direct", "parameters": {"Coef_K_Decom": 0.0057, "Coef_K_Trans": 0.086, "F_notill": 0.93, "F_pH": 0.101, "F_temp": 1.373, "F_texture": 0.899, "F_water": 0.266, "Pool": 281.7}, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling carbon pool respiration loss.\nUse the verified formula:\n\\[R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\\]\nIn calculation-friendly notation, this is:\nRespiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans).\nSubstitute the relevant values:\nPool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction.\nThe computed value is 0.045 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.045}.\n#### 0.045", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "wrong_formula_id": "CMF007"} {"answer": "0.17", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF037", "CMF040"], "effective_parameters": {"PDT": 33.5, "Tmax_day": 38.5}, "family": "growth_yield", "formula": {"expression": "if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1", "id": "CMF035", "latex": "HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}", "name": "高温结实因子", "name_en": "High-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF035_branch_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]", "knowledge_formula_expr": "Formula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise logistic", "mode": "branch", "parameters": {"PDT": 33.5, "Tmax_day": 38.5}, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as High-temperature fertility factor.\nUse the verified formula:\n\\[HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\\]\nIn calculation-friendly notation, this is:\nHTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1.\nSubstitute the relevant values:\nPDT=33.5 physiological days, Tmax_day=38.5 C.\nThe computed value is 0.17 fraction.\nTherefore, the final numeric answer is \\boxed{0.17}.\n#### 0.17", "split": "dev", "subject": "Temperature stress", "template_id": 0, "wrong_formula_id": "CMF037"} {"answer": "0.17", "benchmark_version": "cropmath-v1", "change": {"new_value": 34.8, "old_value": 38.7, "parameter": "PDT"}, "distractor_formula_ids": ["CMF037", "CMF046"], "effective_parameters": {"PDT": 34.8, "Tmax_day": 38.5}, "family": "growth_yield", "formula": {"expression": "if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1", "id": "CMF035", "latex": "HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}", "name": "高温结实因子", "name_en": "High-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF035_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]", "knowledge_formula_expr": "Formula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "parameters": {"PDT": 38.7, "Tmax_day": 38.5}, "problem": "Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as High-temperature fertility factor.\nUse the verified formula:\n\\[HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\\]\nIn calculation-friendly notation, this is:\nHTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1.\nSubstitute the relevant values:\nPDT=34.8 physiological days, Tmax_day=38.5 C.\nThe computed value is 0.17 fraction.\nTherefore, the final numeric answer is \\boxed{0.17}.\n#### 0.17", "split": "dev", "subject": "Temperature stress", "template_id": 1, "wrong_formula_id": "CMF037"} {"answer": "14.11", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF056", "CMF058"], "effective_parameters": {"F_CO2": 1.33, "F_Heat": 0.84, "F_Temp": 0.77, "F_Water": 0.47, "RUE": 1.44, "SRAD": 12.72, "fSolar": 0.16}, "family": "environmental_response", "formula": {"expression": "10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)", "id": "CMF055", "latex": "dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)", "name": "环境响应日生物量增量", "name_en": "environmental-response daily biomass increment", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF055_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]", "knowledge_formula_expr": "Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Multiplicative minimum stress", "mode": "chain_2", "parameters": {"F_CO2": 1.33, "F_Heat": 0.84, "F_Temp": 0.77, "F_Water": 0.47, "RUE": 1.44, "SRAD": 12.72, "fSolar": 0.16}, "problem": "In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily biomass increment.\nUse the verified formula:\n\\[dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\\]\nIn calculation-friendly notation, this is:\ndBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat).\nSubstitute the relevant values:\nRUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction.\nThe computed value is 14.11 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{14.11}.\n#### 14.11", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF056"} {"answer": "0.58", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF051", "CMF042"], "effective_parameters": {"B_above": 12702.0, "PDT": 26.0, "Y_grain": 6715.0}, "family": "growth_yield", "formula": {"expression": "1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))", "id": "CMF045", "latex": "PIS=1-PIGL-PIP", "name": "茎鞘分配指数", "name_en": "Stem partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF045_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]", "knowledge_formula_expr": "Formula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "parameters": {"B_above": 12702.0, "PDT": 26.0, "Y_grain": 6715.0}, "problem": "A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Stem partitioning index.\nUse the verified formula:\n\\[PIS=1-PIGL-PIP\\]\nIn calculation-friendly notation, this is:\nPIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39))).\nSubstitute the relevant values:\nPDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha.\nThe computed value is 0.58 fraction.\nTherefore, the final numeric answer is \\boxed{0.58}.\n#### 0.58", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF051"} {"answer": "0.327", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.0, "old_value": 1.11, "parameter": "SI"}, "distractor_formula_ids": ["CMF019", "CMF023"], "effective_parameters": {"OMN": 23.3, "SI": 1.0, "TI": 0.749, "WI": 0.695}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.027 * OMN", "id": "CMF020", "latex": "OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN", "name": "新鲜有机质碳分解量", "name_en": "Fresh organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF020_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]", "knowledge_formula_expr": "Formula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "parameters": {"OMN": 23.3, "SI": 1.11, "TI": 0.749, "WI": 0.695}, "problem": "Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Fresh organic matter carbon decomposition.\nUse the verified formula:\n\\[OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\\]\nIn calculation-friendly notation, this is:\nOMNC = WI * SI * TI * 0.027 * OMN.\nSubstitute the relevant values:\nWI=0.695 fraction, SI=1.0 fraction, TI=0.749 fraction, OMN=23.3 g C/m2.\nThe computed value is 0.327 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.327}.\n#### 0.327", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF019"} {"answer": "14.86", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.95, "old_value": 1.0, "parameter": "FCO2"}, "distractor_formula_ids": ["CMF047", "CMF029"], "effective_parameters": {"Am": 49.4, "FCO2": 0.95, "FPA": 0.69, "FT": 0.85, "NNI": 0.54, "WDF": 0.66}, "family": "growth_yield", "formula": {"expression": "Am * FCO2 * FPA * FT * min(NNI, WDF)", "id": "CMF026", "latex": "AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)", "name": "实际最大同化速率", "name_en": "Actual maximum assimilation rate", "precision": 2, "unit": "kg CO2/ha/h"}, "id": "CMF026_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_name": "Relevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "parameters": {"Am": 49.4, "FCO2": 1.0, "FPA": 0.69, "FT": 0.85, "NNI": 0.54, "WDF": 0.66}, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual maximum assimilation rate.\nUse the verified formula:\n\\[AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nAMAX = Am * FCO2 * FPA * FT * min(NNI, WDF).\nSubstitute the relevant values:\nAm=49.4 kg CO2/ha/h, FCO2=0.95 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction.\nThe computed value is 14.86 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{14.86}.\n#### 14.86", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF047"} {"answer": "8.687", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF014", "CMF005"], "effective_parameters": {"Day_Length_hour": 15.46, "F_CO2": 0.99, "F_T": 0.39, "F_Water": 0.66, "Photo_Rate_opt": 51.04}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h.\nThe computed value is 8.687 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{8.687}.\n#### 8.687", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF014"} {"answer": "6835.8", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF061", "CMF057"], "effective_parameters": {"Biomass": 12428.7, "HI": 0.55}, "family": "environmental_response", "formula": {"expression": "Biomass * HI", "id": "CMF062", "latex": "Yield=Biomass\\cdot HI", "name": "环境响应产量", "name_en": "environmental-response yield", "precision": 1, "unit": "kg/ha"}, "id": "CMF062_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]", "knowledge_formula_expr": "Formula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha", "knowledge_name": "Relevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"Biomass": 12428.7, "HI": 0.55}, "problem": "Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response yield.\nUse the verified formula:\n\\[Yield=Biomass\\cdot HI\\]\nIn calculation-friendly notation, this is:\nYield = Biomass * HI.\nSubstitute the relevant values:\nBiomass=12428.7 kg/ha, HI=0.55 fraction.\nThe computed value is 6835.8 kg/ha.\nTherefore, the final numeric answer is \\boxed{6835.8}.\n#### 6835.8", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF061"} {"answer": "0.904", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.63, "old_value": 0.55, "parameter": "RPE"}, "distractor_formula_ids": ["CMF029", "CMF037"], "effective_parameters": {"BFF": 0.78, "DTE": 0.904, "IE": 1.23, "PDT": 26.3, "RPE": 0.63}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "parameters": {"BFF": 0.78, "DTE": 0.904, "IE": 1.23, "PDT": 26.3, "RPE": 0.55}, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.63 fraction, BFF=0.78 dimensionless.\nThe computed value is 0.904 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.904}.\n#### 0.904", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF029"} {"answer": "0.500", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF005", "CMF007"], "effective_parameters": {"Coef_K_Decom": 0.0026, "Coef_K_Trans": 0.197, "F_notill": 0.97, "F_pH": 0.43, "F_temp": 0.649, "F_texture": 0.875, "F_water": 0.566, "Pool": 1787.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)", "id": "CMF009", "latex": "R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})", "name": "碳库呼吸损失", "name_en": "carbon-nitrogen cycling carbon pool respiration loss", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF009_chain_2_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]", "knowledge_formula_expr": "Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "parameters": {"Coef_K_Decom": 0.0026, "Coef_K_Trans": 0.197, "F_notill": 0.97, "F_pH": 0.43, "F_temp": 0.649, "F_texture": 0.875, "F_water": 0.566, "Pool": 1787.0}, "problem": "Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling carbon pool respiration loss.\nUse the verified formula:\n\\[R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\\]\nIn calculation-friendly notation, this is:\nRespiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans).\nSubstitute the relevant values:\nPool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction.\nThe computed value is 0.500 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.500}.\n#### 0.500", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "wrong_formula_id": "CMF005"} {"answer": "0.269", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF057", "CMF056"], "effective_parameters": {"ExtremeT": 43.5, "MaxT": 33.1, "Tmax": 40.7}, "family": "environmental_response", "formula": {"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)", "id": "CMF058", "latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}", "name": "环境响应高温响应", "name_en": "environmental-response heat stress response", "precision": 3, "unit": "fraction"}, "id": "CMF058_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]", "knowledge_formula_expr": "Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "parameters": {"ExtremeT": 43.5, "MaxT": 33.1, "Tmax": 40.7}, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response heat stress response.\nUse the verified formula:\n\\[F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\\]\nIn calculation-friendly notation, this is:\nF_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).\nSubstitute the relevant values:\nTmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C.\nThe computed value is 0.269 fraction.\nTherefore, the final numeric answer is \\boxed{0.269}.\n#### 0.269", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF057"} {"answer": "0.37", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.044, "old_value": 0.049, "parameter": "a"}, "distractor_formula_ids": ["CMF027", "CMF028"], "effective_parameters": {"PDT": 50.7, "a": 0.044}, "family": "growth_yield", "formula": {"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))", "id": "CMF030", "latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)", "name": "生理年龄因子", "name_en": "Physiological age factor", "precision": 2, "unit": "fraction"}, "id": "CMF030_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]", "knowledge_formula_expr": "Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "parameters": {"PDT": 50.7, "a": 0.049}, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Physiological age factor.\nUse the verified formula:\n\\[F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\\]\nIn calculation-friendly notation, this is:\nFPA = if PDT<28: 1; else: exp(-a*(PDT-28)).\nSubstitute the relevant values:\nPDT=50.7 physiological days, a=0.044 dimensionless.\nThe computed value is 0.37 fraction.\nTherefore, the final numeric answer is \\boxed{0.37}.\n#### 0.37", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF027"} {"answer": "0.57", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF048", "CMF046"], "effective_parameters": {"B_above": 18531.0, "PDT": 40.8, "Y_grain": 6146.0}, "family": "growth_yield", "formula": {"expression": "1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))", "id": "CMF045", "latex": "PIS=1-PIGL-PIP", "name": "茎鞘分配指数", "name_en": "Stem partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF045_chain_2_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]", "knowledge_formula_expr": "Formula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Residual with piecewise components", "mode": "chain_2", "parameters": {"B_above": 18531.0, "PDT": 40.8, "Y_grain": 6146.0}, "problem": "A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Stem partitioning index.\nUse the verified formula:\n\\[PIS=1-PIGL-PIP\\]\nIn calculation-friendly notation, this is:\nPIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39))).\nSubstitute the relevant values:\nPDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha.\nThe computed value is 0.57 fraction.\nTherefore, the final numeric answer is \\boxed{0.57}.\n#### 0.57", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF048"} {"answer": "128.56", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF048", "CMF043"], "effective_parameters": {"DTGA": 321.4, "Rg": 0.4}, "family": "growth_yield", "formula": {"expression": "Rg * DTGA", "id": "CMF046", "latex": "RG=R_g\\cdot DTGA", "name": "生长呼吸", "name_en": "Growth respiration", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF046_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]", "knowledge_formula_expr": "Formula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"DTGA": 321.4, "Rg": 0.4}, "problem": "A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Growth respiration.\nUse the verified formula:\n\\[RG=R_g\\cdot DTGA\\]\nIn calculation-friendly notation, this is:\nRG = Rg * DTGA.\nSubstitute the relevant values:\nRg=0.4 fraction, DTGA=321.4 kg/ha/day.\nThe computed value is 128.56 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{128.56}.\n#### 128.56", "split": "dev", "subject": "Respiration", "template_id": 0, "wrong_formula_id": "CMF048"} {"answer": "4114.4", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF053", "CMF032"], "effective_parameters": {"AWSP": 4622.9, "grain_fraction": 0.89}, "family": "growth_yield", "formula": {"expression": "AWSP * grain_fraction", "id": "CMF054", "latex": "Y=AWSP\\cdot f_{grain}", "name": "稻谷干重产量", "name_en": "Dry grain yield", "precision": 1, "unit": "kg/ha"}, "id": "CMF054_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Dry grain yield (稻谷干重产量).\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Dry grain yield.\nUse the verified formula:\n\\[Y=AWSP\\cdot f_{grain}\\]\nIn calculation-friendly notation, this is:\nY = AWSP * grain_fraction.\nSubstitute the relevant values:\nAWSP=4622.9 kg/ha, grain_fraction=0.89 fraction.\nThe computed value is 4114.4 kg/ha.\nTherefore, the final numeric answer is \\boxed{4114.4}.\n#### 4114.4", "split": "dev", "subject": "Yield formation", "template_id": 1, "wrong_formula_id": "CMF053"} {"answer": "0.915", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF022", "CMF018"], "effective_parameters": {"Q10": 3.04, "t_soil": 29.2}, "family": "methane_emission", "formula": {"expression": "Q10^((min(t_soil,30)-30)/10)", "id": "CMF023", "latex": "TI=Q_{10}^{(t_{sl}-30)/10}", "name": "甲烷温度指数", "name_en": "Methane temperature index", "precision": 3, "unit": "multiplier"}, "id": "CMF023_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]", "knowledge_formula_expr": "Formula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "parameters": {"Q10": 3.04, "t_soil": 29.2}, "problem": "Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane temperature index.\nUse the verified formula:\n\\[TI=Q_{10}^{(t_{sl}-30)/10}\\]\nIn calculation-friendly notation, this is:\nTI = Q10^((min(t_soil,30)-30)/10).\nSubstitute the relevant values:\nQ10=3.04 dimensionless, t_soil=29.2 C.\nThe computed value is 0.915 fraction.\nTherefore, the final numeric answer is \\boxed{0.915}.\n#### 0.915", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF022"} {"answer": "3886.3", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF055", "CMF058"], "effective_parameters": {"Biomass": 14393.7, "HI": 0.27}, "family": "environmental_response", "formula": {"expression": "Biomass * HI", "id": "CMF062", "latex": "Yield=Biomass\\cdot HI", "name": "环境响应产量", "name_en": "environmental-response yield", "precision": 1, "unit": "kg/ha"}, "id": "CMF062_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]", "knowledge_formula_expr": "Formula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha", "knowledge_name": "Relevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"Biomass": 14393.7, "HI": 0.27}, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response yield.\nUse the verified formula:\n\\[Yield=Biomass\\cdot HI\\]\nIn calculation-friendly notation, this is:\nYield = Biomass * HI.\nSubstitute the relevant values:\nBiomass=14393.7 kg/ha, HI=0.27 fraction.\nThe computed value is 3886.3 kg/ha.\nTherefore, the final numeric answer is \\boxed{3886.3}.\n#### 3886.3", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF055"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": {"new_value": 28.7, "old_value": 25.0, "parameter": "Tmax"}, "distractor_formula_ids": ["CMF055", "CMF057"], "effective_parameters": {"ExtremeT": 47.1, "MaxT": 34.3, "Tmax": 28.7}, "family": "environmental_response", "formula": {"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)", "id": "CMF058", "latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}", "name": "环境响应高温响应", "name_en": "environmental-response heat stress response", "precision": 3, "unit": "fraction"}, "id": "CMF058_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]", "knowledge_formula_expr": "Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "parameters": {"ExtremeT": 47.1, "MaxT": 34.3, "Tmax": 25.0}, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response heat stress response.\nUse the verified formula:\n\\[F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\\]\nIn calculation-friendly notation, this is:\nF_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).\nSubstitute the relevant values:\nTmax=28.7 C, MaxT=34.3 C, ExtremeT=47.1 C.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF055"} {"answer": "0.395", "benchmark_version": "cropmath-v1", "change": {"new_value": 22.6, "old_value": 20.5, "parameter": "t_soil"}, "distractor_formula_ids": ["CMF020", "CMF025"], "effective_parameters": {"Q10": 3.51, "t_soil": 22.6}, "family": "methane_emission", "formula": {"expression": "Q10^((min(t_soil,30)-30)/10)", "id": "CMF023", "latex": "TI=Q_{10}^{(t_{sl}-30)/10}", "name": "甲烷温度指数", "name_en": "Methane temperature index", "precision": 3, "unit": "multiplier"}, "id": "CMF023_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]", "knowledge_formula_expr": "Formula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "parameters": {"Q10": 3.51, "t_soil": 20.5}, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane temperature index.\nUse the verified formula:\n\\[TI=Q_{10}^{(t_{sl}-30)/10}\\]\nIn calculation-friendly notation, this is:\nTI = Q10^((min(t_soil,30)-30)/10).\nSubstitute the relevant values:\nQ10=3.51 dimensionless, t_soil=22.6 C.\nThe computed value is 0.395 fraction.\nTherefore, the final numeric answer is \\boxed{0.395}.\n#### 0.395", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF020"} {"answer": "3.432", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF012", "CMF010"], "effective_parameters": {"Day_Length_hour": 10.37, "F_CO2": 1.28, "F_T": 0.49, "F_Water": 0.33, "Photo_Rate_opt": 37.01}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_chain_2_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]", "knowledge_formula_expr": "Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "parameters": {"Day_Length_hour": 10.37, "F_CO2": 1.28, "F_T": 0.49, "F_Water": 0.33, "Photo_Rate_opt": 37.01}, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h.\nThe computed value is 3.432 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{3.432}.\n#### 3.432", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF012"} {"answer": "17.02", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF053", "CMF028"], "effective_parameters": {"Am": 31.1, "FCO2": 1.2, "FPA": 0.92, "FT": 0.67, "NNI": 0.74, "WDF": 0.92}, "family": "growth_yield", "formula": {"expression": "Am * FCO2 * FPA * FT * min(NNI, WDF)", "id": "CMF026", "latex": "AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)", "name": "实际最大同化速率", "name_en": "Actual maximum assimilation rate", "precision": 2, "unit": "kg CO2/ha/h"}, "id": "CMF026_chain_2_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_name": "Relevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "parameters": {"Am": 31.1, "FCO2": 1.2, "FPA": 0.92, "FT": 0.67, "NNI": 0.74, "WDF": 0.92}, "problem": "Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual maximum assimilation rate.\nUse the verified formula:\n\\[AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nAMAX = Am * FCO2 * FPA * FT * min(NNI, WDF).\nSubstitute the relevant values:\nAm=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction.\nThe computed value is 17.02 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{17.02}.\n#### 17.02", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF053"} {"answer": "1.310", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.68, "old_value": 0.75, "parameter": "Coef_B_CO2"}, "distractor_formula_ids": ["CMF002", "CMF013"], "effective_parameters": {"CO2": 536.0, "Coef_B_CO2": 0.68}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "1 + Coef_B_CO2 * ln(CO2 / 340)", "id": "CMF001", "latex": "F_{CO2}=1+B\\ln(CO2/340)", "name": "作物二氧化碳施肥因子", "name_en": "carbon-nitrogen cycling CO2 fertilization factor", "precision": 3, "unit": "multiplier"}, "id": "CMF001_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]", "knowledge_formula_expr": "Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "parameters": {"CO2": 536.0, "Coef_B_CO2": 0.75}, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling CO2 fertilization factor.\nUse the verified formula:\n\\[F_{CO2}=1+B\\ln(CO2/340)\\]\nIn calculation-friendly notation, this is:\nF_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340).\nSubstitute the relevant values:\nCO2=536.0 ppm, Coef_B_CO2=0.68 dimensionless.\nThe computed value is 1.310 fraction.\nTherefore, the final numeric answer is \\boxed{1.310}.\n#### 1.310", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF002"} {"answer": "0.94", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF050", "CMF026"], "effective_parameters": {"PDT": 28.5, "Qt": 6.2, "Tmean": 16.2}, "family": "growth_yield", "formula": {"expression": "if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1", "id": "CMF039", "latex": "LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}", "name": "低温结实因子", "name_en": "Low-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF039_branch_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]", "knowledge_formula_expr": "Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "parameters": {"PDT": 28.5, "Qt": 6.2, "Tmean": 16.2}, "problem": "For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Low-temperature fertility factor.\nUse the verified formula:\n\\[LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\\]\nIn calculation-friendly notation, this is:\nLTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1.\nSubstitute the relevant values:\nPDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day.\nThe computed value is 0.94 fraction.\nTherefore, the final numeric answer is \\boxed{0.94}.\n#### 0.94", "split": "dev", "subject": "Temperature stress", "template_id": 1, "wrong_formula_id": "CMF050"} {"answer": "0.900", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF008"], "effective_parameters": {"Coef_T_M": 0.16, "Td": 19.3, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)", "id": "CMF013", "latex": "F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}", "name": "作物温度活性因子", "name_en": "carbon-nitrogen cycling crop temperature activity factor", "precision": 3, "unit": "response factor"}, "id": "CMF013_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]", "knowledge_formula_expr": "Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Beta temperature response", "mode": "direct", "parameters": {"Coef_T_M": 0.16, "Td": 19.3, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "problem": "A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop temperature activity factor.\nUse the verified formula:\n\\[F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\\]\nIn calculation-friendly notation, this is:\nF_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M).\nSubstitute the relevant values:\nTd=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless.\nThe computed value is 0.900 fraction.\nTherefore, the final numeric answer is \\boxed{0.900}.\n#### 0.900", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 0, "wrong_formula_id": "CMF009"} {"answer": "0.95", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF026", "CMF046"], "effective_parameters": {"PDT": 30.8, "Qt": 4.3, "Tmean": 20.6}, "family": "growth_yield", "formula": {"expression": "if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1", "id": "CMF039", "latex": "LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}", "name": "低温结实因子", "name_en": "Low-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF039_branch_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]", "knowledge_formula_expr": "Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "parameters": {"PDT": 30.8, "Qt": 4.3, "Tmean": 20.6}, "problem": "Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Low-temperature fertility factor.\nUse the verified formula:\n\\[LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\\]\nIn calculation-friendly notation, this is:\nLTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1.\nSubstitute the relevant values:\nPDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day.\nThe computed value is 0.95 fraction.\nTherefore, the final numeric answer is \\boxed{0.95}.\n#### 0.95", "split": "dev", "subject": "Temperature stress", "template_id": 2, "wrong_formula_id": "CMF026"} {"answer": "3.943", "benchmark_version": "cropmath-v1", "change": {"new_value": 10.12, "old_value": 11.24, "parameter": "Day_Length_hour"}, "distractor_formula_ids": ["CMF001", "CMF007"], "effective_parameters": {"Day_Length_hour": 10.12, "F_CO2": 1.15, "F_T": 0.64, "F_Water": 0.44, "Photo_Rate_opt": 27.85}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]", "knowledge_formula_expr": "Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "parameters": {"Day_Length_hour": 11.24, "F_CO2": 1.15, "F_T": 0.64, "F_Water": 0.44, "Photo_Rate_opt": 27.85}, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=10.12 h.\nThe computed value is 3.943 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{3.943}.\n#### 3.943", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF001"} {"answer": "0.95", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.6, "old_value": 1.8, "parameter": "Qt"}, "distractor_formula_ids": ["CMF029", "CMF030"], "effective_parameters": {"PDT": 35.6, "Qt": 1.6, "Tmean": 18.5}, "family": "growth_yield", "formula": {"expression": "if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1", "id": "CMF039", "latex": "LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}", "name": "低温结实因子", "name_en": "Low-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF039_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]", "knowledge_formula_expr": "Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "parameters": {"PDT": 35.6, "Qt": 1.8, "Tmean": 18.5}, "problem": "Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Low-temperature fertility factor.\nUse the verified formula:\n\\[LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\\]\nIn calculation-friendly notation, this is:\nLTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1.\nSubstitute the relevant values:\nPDT=35.6 physiological days, Tmean=18.5 C, Qt=1.6 C day.\nThe computed value is 0.95 fraction.\nTherefore, the final numeric answer is \\boxed{0.95}.\n#### 0.95", "split": "dev", "subject": "Temperature stress", "template_id": 1, "wrong_formula_id": "CMF029"} {"answer": "0.166", "benchmark_version": "cropmath-v1", "change": {"new_value": 2200.0, "old_value": 2121.5, "parameter": "Pool"}, "distractor_formula_ids": ["CMF001", "CMF014"], "effective_parameters": {"Coef_K_Decom": 0.001, "Coef_K_Trans": 0.264, "F_notill": 0.9, "F_pH": 0.409, "F_temp": 0.838, "F_texture": 0.77, "F_water": 0.432, "Pool": 2200.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)", "id": "CMF009", "latex": "R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})", "name": "碳库呼吸损失", "name_en": "carbon-nitrogen cycling carbon pool respiration loss", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF009_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling carbon pool respiration loss.\nUse the verified formula:\n\\[R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\\]\nIn calculation-friendly notation, this is:\nRespiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans).\nSubstitute the relevant values:\nPool=2200.0 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction.\nThe computed value is 0.166 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.166}.\n#### 0.166", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "wrong_formula_id": "CMF001"} {"answer": "27.35", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF025", "CMF020"], "effective_parameters": {"Tair": 30.2}, "family": "methane_emission", "formula": {"expression": "4.4 + 0.76 * Tair", "id": "CMF024", "latex": "T_{soil}=4.4+0.76T_{air}", "name": "土壤温度估算", "name_en": "Soil temperature from air temperature", "precision": 2, "unit": "C"}, "id": "CMF024_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]", "knowledge_formula_expr": "Formula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_name": "Relevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear", "mode": "direct", "parameters": {"Tair": 30.2}, "problem": "Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil temperature from air temperature.\nUse the verified formula:\n\\[T_{soil}=4.4+0.76T_{air}\\]\nIn calculation-friendly notation, this is:\nTsoil = 4.4 + 0.76 * Tair.\nSubstitute the relevant values:\nTair=30.2 C.\nThe computed value is 27.35 C.\nTherefore, the final numeric answer is \\boxed{27.35}.\n#### 27.35", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF025"} {"answer": "12.55", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF032", "CMF035"], "effective_parameters": {"PS": 0.237, "Po": 10.5}, "family": "growth_yield", "formula": {"expression": "Po + sqrt(1 / PS)", "id": "CMF040", "latex": "P_c=P_o+\\sqrt{1/PS}", "name": "临界日长", "name_en": "Critical daylength", "precision": 2, "unit": "h"}, "id": "CMF040_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]", "knowledge_formula_expr": "Formula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_name": "Relevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Critical daylength.\nUse the verified formula:\n\\[P_c=P_o+\\sqrt{1/PS}\\]\nIn calculation-friendly notation, this is:\nPc = Po + sqrt(1 / PS).\nSubstitute the relevant values:\nPo=10.5 h, PS=0.237 dimensionless.\nThe computed value is 12.55 h.\nTherefore, the final numeric answer is \\boxed{12.55}.\n#### 12.55", "split": "dev", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF032"} {"answer": "275.26", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF039", "CMF034"], "effective_parameters": {"DTGA": 407.9, "RG": 50.3, "RM": 96.1, "b": 0.05}, "family": "growth_yield", "formula": {"expression": "(DTGA - RM - RG) / (1 - b)", "id": "CMF033", "latex": "GCR=\\frac{DTGA-RM-RG}{1-b}", "name": "日干物质增量", "name_en": "Daily dry matter increment", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF033_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]", "knowledge_formula_expr": "Formula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "parameters": {"DTGA": 407.9, "RG": 50.3, "RM": 96.1, "b": 0.05}, "problem": "Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily dry matter increment.\nUse the verified formula:\n\\[GCR=\\frac{DTGA-RM-RG}{1-b}\\]\nIn calculation-friendly notation, this is:\nGCR = (DTGA - RM - RG) / (1 - b).\nSubstitute the relevant values:\nDTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction.\nThe computed value is 275.26 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{275.26}.\n#### 275.26", "split": "dev", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF039"} {"answer": "1.518", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF010", "CMF011"], "effective_parameters": {"Coef_T_M": 0.76, "Td": 33.8, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)", "id": "CMF013", "latex": "F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}", "name": "作物温度活性因子", "name_en": "carbon-nitrogen cycling crop temperature activity factor", "precision": 3, "unit": "response factor"}, "id": "CMF013_branch_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]", "knowledge_formula_expr": "Formula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Beta temperature response", "mode": "branch", "parameters": {"Coef_T_M": 0.76, "Td": 33.8, "Tmax_active": 42.0, "Tmin_active": 3.0, "Topt_active": 25.0}, "problem": "Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop temperature activity factor.\nUse the verified formula:\n\\[F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\\]\nIn calculation-friendly notation, this is:\nF_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M).\nSubstitute the relevant values:\nTd=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless.\nThe computed value is 1.518 fraction.\nTherefore, the final numeric answer is \\boxed{1.518}.\n#### 1.518", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "wrong_formula_id": "CMF010"} {"answer": "9.179", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF022", "CMF023"], "effective_parameters": {"CH4RiceEfC": 0.386, "Ebl": 4.01, "P": 15.119, "W": 4936.9, "Wmax": 12837.1}, "family": "methane_emission", "formula": {"expression": "Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25", "id": "CMF017", "latex": "E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}", "name": "甲烷总排放", "name_en": "Total methane emission", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF017_chain_2_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]", "knowledge_formula_expr": "Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Chained sum", "mode": "chain_2", "parameters": {"CH4RiceEfC": 0.386, "Ebl": 4.01, "P": 15.119, "W": 4936.9, "Wmax": 12837.1}, "problem": "Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Total methane emission.\nUse the verified formula:\n\\[E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\\]\nIn calculation-friendly notation, this is:\nE = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25.\nSubstitute the relevant values:\nEbl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha.\nThe computed value is 9.179 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{9.179}.\n#### 9.179", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF022"} {"answer": "0.629", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF004"], "effective_parameters": {"Biom_plant_yesterday": 221.0, "Coef_Rm": 0.0016, "DVI": 1.15, "Tmn": 25.3}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "parameters": {"Biom_plant_yesterday": 221.0, "Coef_Rm": 0.0016, "DVI": 1.15, "Tmn": 25.3}, "problem": "In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless.\nThe computed value is 0.629 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.629}.\n#### 0.629", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "wrong_formula_id": "CMF009"} {"answer": "12.27", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF035", "CMF052"], "effective_parameters": {"PS": 0.318, "Po": 10.5}, "family": "growth_yield", "formula": {"expression": "Po + sqrt(1 / PS)", "id": "CMF040", "latex": "P_c=P_o+\\sqrt{1/PS}", "name": "临界日长", "name_en": "Critical daylength", "precision": 2, "unit": "h"}, "id": "CMF040_chain_2_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]", "knowledge_formula_expr": "Formula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_name": "Relevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Power response", "mode": "chain_2", "parameters": {"PS": 0.318, "Po": 10.5}, "problem": "A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Critical daylength.\nUse the verified formula:\n\\[P_c=P_o+\\sqrt{1/PS}\\]\nIn calculation-friendly notation, this is:\nPc = Po + sqrt(1 / PS).\nSubstitute the relevant values:\nPo=10.5 h, PS=0.318 dimensionless.\nThe computed value is 12.27 h.\nTherefore, the final numeric answer is \\boxed{12.27}.\n#### 12.27", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF035"} {"answer": "0.825", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF016", "CMF020"], "effective_parameters": {"Eh": -133.0}, "family": "methane_emission", "formula": {"expression": "if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))", "id": "CMF018", "latex": "F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}", "name": "氧化还原电位因子", "name_en": "Redox potential factor", "precision": 3, "unit": "fraction"}, "id": "CMF018_branch_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]", "knowledge_formula_expr": "Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "parameters": {"Eh": -133.0}, "problem": "A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Redox potential factor.\nUse the verified formula:\n\\[F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\\]\nIn calculation-friendly notation, this is:\nFEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150)).\nSubstitute the relevant values:\nEh=-133.0 mV.\nThe computed value is 0.825 fraction.\nTherefore, the final numeric answer is \\boxed{0.825}.\n#### 0.825", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF016"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF059", "CMF057"], "effective_parameters": {"ExtremeT": 42.5, "MaxT": 35.1, "Tmax": 33.9}, "family": "environmental_response", "formula": {"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)", "id": "CMF058", "latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}", "name": "环境响应高温响应", "name_en": "environmental-response heat stress response", "precision": 3, "unit": "fraction"}, "id": "CMF058_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]", "knowledge_formula_expr": "Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "parameters": {"ExtremeT": 42.5, "MaxT": 35.1, "Tmax": 33.9}, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response heat stress response.\nUse the verified formula:\n\\[F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\\]\nIn calculation-friendly notation, this is:\nF_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).\nSubstitute the relevant values:\nTmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF059"} {"answer": "77.039", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.9, "old_value": 0.78, "parameter": "SI"}, "distractor_formula_ids": ["CMF021", "CMF022"], "effective_parameters": {"SI": 0.9, "VI": 1.0, "W": 5517.7}, "family": "methane_emission", "formula": {"expression": "0.0018 * VI * SI * W ** 1.25", "id": "CMF015", "latex": "C_r=0.0018VI\\cdot SI\\cdot W^{1.25}", "name": "根系分泌碳底物", "name_en": "Root-exudate carbon substrate", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF015_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]", "knowledge_formula_expr": "Formula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Root-exudate carbon substrate (根系分泌碳底物).\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Power product", "mode": "sensitivity", "parameters": {"SI": 0.78, "VI": 1.0, "W": 5517.7}, "problem": "Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root-exudate carbon substrate (根系分泌碳底物).\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root-exudate carbon substrate.\nUse the verified formula:\n\\[C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\\]\nIn calculation-friendly notation, this is:\nCr = 0.0018 * VI * SI * W ** 1.25.\nSubstitute the relevant values:\nVI=1.0 fraction, SI=0.9 fraction, W=5517.7 kg/ha.\nThe computed value is 77.039 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{77.039}.\n#### 77.039", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF021"} {"answer": "0.09", "benchmark_version": "cropmath-v1", "change": {"new_value": 32.7, "old_value": 29.7, "parameter": "PDT"}, "distractor_formula_ids": ["CMF054", "CMF031"], "effective_parameters": {"B_above": 12138.0, "PDT": 32.7, "Y_grain": 6390.0}, "family": "growth_yield", "formula": {"expression": "(Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))", "id": "CMF042", "latex": "PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}", "name": "穗分配指数", "name_en": "Panicle partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF042_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]", "knowledge_formula_expr": "Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multi-step sigmoid", "mode": "sensitivity", "parameters": {"B_above": 12138.0, "PDT": 29.7, "Y_grain": 6390.0}, "problem": "Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Panicle partitioning index.\nUse the verified formula:\n\\[PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\\]\nIn calculation-friendly notation, this is:\nPIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39))).\nSubstitute the relevant values:\nY_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=32.7 physiological days.\nThe computed value is 0.09 fraction.\nTherefore, the final numeric answer is \\boxed{0.09}.\n#### 0.09", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF054"} {"answer": "0.06", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF053", "CMF038"], "effective_parameters": {"PDT": 38.5, "Tmax_day": 39.8}, "family": "growth_yield", "formula": {"expression": "if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1", "id": "CMF035", "latex": "HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}", "name": "高温结实因子", "name_en": "High-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF035_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]", "knowledge_formula_expr": "Formula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise logistic", "mode": "direct", "parameters": {"PDT": 38.5, "Tmax_day": 39.8}, "problem": "Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as High-temperature fertility factor.\nUse the verified formula:\n\\[HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\\]\nIn calculation-friendly notation, this is:\nHTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1.\nSubstitute the relevant values:\nPDT=38.5 physiological days, Tmax_day=39.8 C.\nThe computed value is 0.06 fraction.\nTherefore, the final numeric answer is \\boxed{0.06}.\n#### 0.06", "split": "dev", "subject": "Temperature stress", "template_id": 1, "wrong_formula_id": "CMF053"} {"answer": "0.92", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF028", "CMF027"], "effective_parameters": {"AWL": 7061.0, "SLA": 0.00013}, "family": "growth_yield", "formula": {"expression": "SLA * AWL", "id": "CMF038", "latex": "LAI=SLA\\cdot AWL", "name": "比叶面积法叶面积指数", "name_en": "Leaf area index from SLA", "precision": 2, "unit": "m2/m2"}, "id": "CMF038_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]", "knowledge_formula_expr": "Formula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_name": "Relevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"AWL": 7061.0, "SLA": 0.00013}, "problem": "Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Leaf area index from SLA.\nUse the verified formula:\n\\[LAI=SLA\\cdot AWL\\]\nIn calculation-friendly notation, this is:\nLAI = SLA * AWL.\nSubstitute the relevant values:\nSLA=0.00013 ha/kg, AWL=7061.0 kg/ha.\nThe computed value is 0.92 m2/m2.\nTherefore, the final numeric answer is \\boxed{0.92}.\n#### 0.92", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF028"} {"answer": "18.864", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF006", "CMF009"], "effective_parameters": {"Coef_alfa": 16.16, "Coef_beta": 129.6, "Leaf_N": 1.36, "PAR": 785.1}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "(Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)", "id": "CMF007", "latex": "P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}", "name": "最适光合速率", "name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "precision": 3, "unit": "umol CO2/m2/s"}, "id": "CMF007_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]", "knowledge_formula_expr": "Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "parameters": {"Coef_alfa": 16.16, "Coef_beta": 129.6, "Leaf_N": 1.36, "PAR": 785.1}, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling optimal photosynthetic rate.\nUse the verified formula:\n\\[P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\\]\nIn calculation-friendly notation, this is:\nPhoto_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta).\nSubstitute the relevant values:\nCoef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2.\nThe computed value is 18.864 umol CO2/m2/s.\nTherefore, the final numeric answer is \\boxed{18.864}.\n#### 18.864", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF006"} {"answer": "5.811", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF002", "CMF014"], "effective_parameters": {"Temp": 30.1, "is_soil_temp": 1.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)", "id": "CMF011", "latex": "F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T", "name": "土壤碳分解温度因子", "name_en": "carbon-nitrogen cycling soil carbon temperature factor", "precision": 3, "unit": "multiplier"}, "id": "CMF011_branch_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]", "knowledge_formula_expr": "Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "parameters": {"Temp": 30.1, "is_soil_temp": 1.0}, "problem": "Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil carbon temperature factor.\nUse the verified formula:\n\\[F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\\]\nIn calculation-friendly notation, this is:\nF_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1).\nSubstitute the relevant values:\nTemp=30.1 C, is_soil_temp=1.0 0/1.\nThe computed value is 5.811 fraction.\nTherefore, the final numeric answer is \\boxed{5.811}.\n#### 5.811", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "wrong_formula_id": "CMF002"} {"answer": "0.290", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF030", "CMF048"], "effective_parameters": {"BFF": 0.86, "DTE": 0.337, "IE": 0.73, "PDT": 51.3, "RPE": 0.26}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "parameters": {"BFF": 0.86, "DTE": 0.337, "IE": 0.73, "PDT": 51.3, "RPE": 0.26}, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless.\nThe computed value is 0.290 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.290}.\n#### 0.290", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF030"} {"answer": "0.758", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.579, "old_value": 0.643, "parameter": "WaterC"}, "distractor_formula_ids": ["CMF023", "CMF015"], "effective_parameters": {"WaterC": 0.579}, "family": "methane_emission", "formula": {"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)", "id": "CMF025", "latex": "WI=0.49e^{3.88W_c-5.4W_c^2}", "name": "水分分解因子", "name_en": "Water-content decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF025_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]", "knowledge_formula_expr": "Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "parameters": {"WaterC": 0.643}, "problem": "Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Water-content decomposition factor.\nUse the verified formula:\n\\[WI=0.49e^{3.88W_c-5.4W_c^2}\\]\nIn calculation-friendly notation, this is:\nWI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).\nSubstitute the relevant values:\nWaterC=0.579 fraction.\nThe computed value is 0.758 fraction.\nTherefore, the final numeric answer is \\boxed{0.758}.\n#### 0.758", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF023"} {"answer": "0.502", "benchmark_version": "cropmath-v1", "change": {"new_value": 32.2, "old_value": 29.3, "parameter": "PDT"}, "distractor_formula_ids": ["CMF026", "CMF048"], "effective_parameters": {"PDT": 32.2}, "family": "growth_yield", "formula": {"expression": "0.0087 * PDT + 0.2222", "id": "CMF036", "latex": "K=0.0087PDT+0.2222", "name": "冠层消光系数", "name_en": "Canopy extinction coefficient", "precision": 3, "unit": "dimensionless"}, "id": "CMF036_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless", "knowledge_name": "Relevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Linear", "mode": "sensitivity", "parameters": {"PDT": 29.3}, "problem": "In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Canopy extinction coefficient.\nUse the verified formula:\n\\[K=0.0087PDT+0.2222\\]\nIn calculation-friendly notation, this is:\nK = 0.0087 * PDT + 0.2222.\nSubstitute the relevant values:\nPDT=32.2 physiological days.\nThe computed value is 0.502 dimensionless.\nTherefore, the final numeric answer is \\boxed{0.502}.\n#### 0.502", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF026"} {"answer": "0.654", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF037", "CMF043"], "effective_parameters": {"BFF": 0.89, "DTE": 0.735, "IE": 0.8, "PDT": 46.1, "RPE": 0.7}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_branch_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "parameters": {"BFF": 0.89, "DTE": 0.735, "IE": 0.8, "PDT": 46.1, "RPE": 0.7}, "problem": "Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless.\nThe computed value is 0.654 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.654}.\n#### 0.654", "split": "dev", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF037"} {"answer": "19.87", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF029", "CMF050"], "effective_parameters": {"GDD": 774.4, "SLAc": 146.5}, "family": "growth_yield", "formula": {"expression": "if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)", "id": "CMF053", "latex": "SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)", "name": "比叶面积", "name_en": "Specific leaf area", "precision": 2, "unit": "cm2/g"}, "id": "CMF053_branch_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]", "knowledge_formula_expr": "Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_name": "Relevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "parameters": {"GDD": 774.4, "SLAc": 146.5}, "problem": "Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Specific leaf area.\nUse the verified formula:\n\\[SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\\]\nIn calculation-friendly notation, this is:\nSLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200).\nSubstitute the relevant values:\nGDD=774.4 C day, SLAc=146.5 cm2/g.\nThe computed value is 19.87 cm2/g.\nTherefore, the final numeric answer is \\boxed{19.87}.\n#### 19.87", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF029"} {"answer": "0.976", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF017", "CMF022"], "effective_parameters": {"WaterC": 0.321}, "family": "methane_emission", "formula": {"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)", "id": "CMF025", "latex": "WI=0.49e^{3.88W_c-5.4W_c^2}", "name": "水分分解因子", "name_en": "Water-content decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF025_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]", "knowledge_formula_expr": "Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "parameters": {"WaterC": 0.321}, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Water-content decomposition factor.\nUse the verified formula:\n\\[WI=0.49e^{3.88W_c-5.4W_c^2}\\]\nIn calculation-friendly notation, this is:\nWI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).\nSubstitute the relevant values:\nWaterC=0.321 fraction.\nThe computed value is 0.976 fraction.\nTherefore, the final numeric answer is \\boxed{0.976}.\n#### 0.976", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF017"} {"answer": "75.74", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF061", "CMF058"], "effective_parameters": {"F_CO2": 1.19, "F_Heat": 0.45, "F_Temp": 0.75, "F_Water": 0.8, "RUE": 1.99, "SRAD": 24.3, "fSolar": 0.39}, "family": "environmental_response", "formula": {"expression": "10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)", "id": "CMF055", "latex": "dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)", "name": "环境响应日生物量增量", "name_en": "environmental-response daily biomass increment", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF055_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]", "knowledge_formula_expr": "Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "parameters": {"F_CO2": 1.19, "F_Heat": 0.45, "F_Temp": 0.75, "F_Water": 0.8, "RUE": 1.99, "SRAD": 24.3, "fSolar": 0.39}, "problem": "In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily biomass increment.\nUse the verified formula:\n\\[dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\\]\nIn calculation-friendly notation, this is:\ndBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat).\nSubstitute the relevant values:\nRUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction.\nThe computed value is 75.74 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{75.74}.\n#### 75.74", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF061"} {"answer": "14.89", "benchmark_version": "cropmath-v1", "change": {"new_value": 13.8, "old_value": 12.0, "parameter": "Tair"}, "distractor_formula_ids": ["CMF015", "CMF018"], "effective_parameters": {"Tair": 13.8}, "family": "methane_emission", "formula": {"expression": "4.4 + 0.76 * Tair", "id": "CMF024", "latex": "T_{soil}=4.4+0.76T_{air}", "name": "土壤温度估算", "name_en": "Soil temperature from air temperature", "precision": 2, "unit": "C"}, "id": "CMF024_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]", "knowledge_formula_expr": "Formula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_name": "Relevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Linear", "mode": "sensitivity", "parameters": {"Tair": 12.0}, "problem": "In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil temperature from air temperature.\nUse the verified formula:\n\\[T_{soil}=4.4+0.76T_{air}\\]\nIn calculation-friendly notation, this is:\nTsoil = 4.4 + 0.76 * Tair.\nSubstitute the relevant values:\nTair=13.8 C.\nThe computed value is 14.89 C.\nTherefore, the final numeric answer is \\boxed{14.89}.\n#### 14.89", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF015"} {"answer": "254.39", "benchmark_version": "cropmath-v1", "change": {"new_value": 5.23, "old_value": 5.81, "parameter": "LAI"}, "distractor_formula_ids": ["CMF005", "CMF011"], "effective_parameters": {"Coef_K": 0.71, "Coef_rou": 0.18, "LAI": 5.23, "Radi": 583.4}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)", "id": "CMF006", "latex": "PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}", "name": "冠层吸收光合有效辐射", "name_en": "carbon-nitrogen cycling canopy absorbed PAR", "precision": 2, "unit": "umol/m2/s"}, "id": "CMF006_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]", "knowledge_formula_expr": "Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "parameters": {"Coef_K": 0.71, "Coef_rou": 0.18, "LAI": 5.81, "Radi": 583.4}, "problem": "Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling canopy absorbed PAR.\nUse the verified formula:\n\\[PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\\]\nIn calculation-friendly notation, this is:\nPAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI).\nSubstitute the relevant values:\nRadi=583.4 W/m2, LAI=5.23 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless.\nThe computed value is 254.39 umol/m2/s.\nTherefore, the final numeric answer is \\boxed{254.39}.\n#### 254.39", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "wrong_formula_id": "CMF005"} {"answer": "5.576", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF010", "CMF011"], "effective_parameters": {"Biom_plant_yesterday": 641.5, "Coef_Rm": 0.0033, "DVI": 1.12, "Tmn": 30.2}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_branch_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "parameters": {"Biom_plant_yesterday": 641.5, "Coef_Rm": 0.0033, "DVI": 1.12, "Tmn": 30.2}, "problem": "A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless.\nThe computed value is 5.576 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{5.576}.\n#### 5.576", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "wrong_formula_id": "CMF010"} {"answer": "0.433", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF015", "CMF020"], "effective_parameters": {"Com": 4.036, "Cr": 10.96, "FEh": 0.128, "TI": 0.774}, "family": "methane_emission", "formula": {"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))", "id": "CMF022", "latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))", "name": "甲烷产生量", "name_en": "Methane production", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF022_chain_2_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]", "knowledge_formula_expr": "Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "parameters": {"Com": 4.036, "Cr": 10.96, "FEh": 0.128, "TI": 0.774}, "problem": "A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane production.\nUse the verified formula:\n\\[P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\\]\nIn calculation-friendly notation, this is:\nP = max(0, 0.27 * FEh * (TI * Cr + Com)).\nSubstitute the relevant values:\nFEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day.\nThe computed value is 0.433 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.433}.\n#### 0.433", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF015"} {"answer": "0.90", "benchmark_version": "cropmath-v1", "change": {"new_value": 42.4, "old_value": 47.1, "parameter": "PDT"}, "distractor_formula_ids": ["CMF038", "CMF040"], "effective_parameters": {"PDT": 42.4}, "family": "growth_yield", "formula": {"expression": "-0.0000842 * PDT^2 + 0.01 * PDT + 0.63", "id": "CMF044", "latex": "PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63", "name": "地上部分配指数", "name_en": "Shoot partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF044_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Shoot partitioning index (地上部分配指数).\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "parameters": {"PDT": 47.1}, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Shoot partitioning index (地上部分配指数).\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Shoot partitioning index.\nUse the verified formula:\n\\[PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\\]\nIn calculation-friendly notation, this is:\nPISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63.\nSubstitute the relevant values:\nPDT=42.4 physiological days.\nThe computed value is 0.90 fraction.\nTherefore, the final numeric answer is \\boxed{0.90}.\n#### 0.90", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF038"} {"answer": "4.656", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF012", "CMF014"], "effective_parameters": {"Biom_plant_yesterday": 757.9, "Coef_Rm": 0.003, "DVI": 0.76, "Tmn": 23.3}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_branch_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "parameters": {"Biom_plant_yesterday": 757.9, "Coef_Rm": 0.003, "DVI": 0.76, "Tmn": 23.3}, "problem": "For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless.\nThe computed value is 4.656 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{4.656}.\n#### 4.656", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "wrong_formula_id": "CMF012"} {"answer": "2.90", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF061", "CMF059"], "effective_parameters": {"Tbase": 8.2, "Tmean": 11.1}, "family": "environmental_response", "formula": {"expression": "max(Tmean - Tbase, 0)", "id": "CMF056", "latex": "dTT=\\max(T_{mean}-T_{base},0)", "name": "每日有效积温", "name_en": "environmental-response daily thermal time", "precision": 2, "unit": "C day"}, "id": "CMF056_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]", "knowledge_formula_expr": "Formula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Maximum linear", "mode": "direct", "parameters": {"Tbase": 8.2, "Tmean": 11.1}, "problem": "Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily thermal time.\nUse the verified formula:\n\\[dTT=\\max(T_{mean}-T_{base},0)\\]\nIn calculation-friendly notation, this is:\ndTT = max(Tmean - Tbase, 0).\nSubstitute the relevant values:\nTmean=11.1 C, Tbase=8.2 C.\nThe computed value is 2.90 C day.\nTherefore, the final numeric answer is \\boxed{2.90}.\n#### 2.90", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF061"} {"answer": "3.15", "benchmark_version": "cropmath-v1", "change": {"new_value": 8500.0, "old_value": 8345.0, "parameter": "AWL"}, "distractor_formula_ids": ["CMF041", "CMF053"], "effective_parameters": {"AWL": 8500.0, "SLA": 0.00037}, "family": "growth_yield", "formula": {"expression": "SLA * AWL", "id": "CMF038", "latex": "LAI=SLA\\cdot AWL", "name": "比叶面积法叶面积指数", "name_en": "Leaf area index from SLA", "precision": 2, "unit": "m2/m2"}, "id": "CMF038_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]", "knowledge_formula_expr": "Formula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_name": "Relevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product", "mode": "sensitivity", "parameters": {"AWL": 8345.0, "SLA": 0.00037}, "problem": "In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Leaf area index from SLA.\nUse the verified formula:\n\\[LAI=SLA\\cdot AWL\\]\nIn calculation-friendly notation, this is:\nLAI = SLA * AWL.\nSubstitute the relevant values:\nSLA=0.00037 ha/kg, AWL=8500.0 kg/ha.\nThe computed value is 3.15 m2/m2.\nTherefore, the final numeric answer is \\boxed{3.15}.\n#### 3.15", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF041"} {"answer": "0.0065", "benchmark_version": "cropmath-v1", "change": {"new_value": 25.9, "old_value": 28.8, "parameter": "PDT"}, "distractor_formula_ids": ["CMF030", "CMF027"], "effective_parameters": {"PDT": 25.9}, "family": "growth_yield", "formula": {"expression": "0.0091 - 0.0001 * PDT", "id": "CMF049", "latex": "RM(T_o)=0.0091-0.0001PDT", "name": "最适温度维持呼吸系数", "name_en": "Maintenance respiration coefficient at optimum temperature", "precision": 4, "unit": "d-1"}, "id": "CMF049_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_name": "Relevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Linear decline", "mode": "sensitivity", "parameters": {"PDT": 28.8}, "problem": "A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration coefficient at optimum temperature.\nUse the verified formula:\n\\[RM(T_o)=0.0091-0.0001PDT\\]\nIn calculation-friendly notation, this is:\nRMTo = 0.0091 - 0.0001 * PDT.\nSubstitute the relevant values:\nPDT=25.9 physiological days.\nThe computed value is 0.0065 d-1.\nTherefore, the final numeric answer is \\boxed{0.0065}.\n#### 0.0065", "split": "dev", "subject": "Respiration", "template_id": 0, "wrong_formula_id": "CMF030"} {"answer": "0.652", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF016", "CMF025"], "effective_parameters": {"OMN": 195.9, "SI": 0.65, "TI": 0.68, "WI": 0.279}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.027 * OMN", "id": "CMF020", "latex": "OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN", "name": "新鲜有机质碳分解量", "name_en": "Fresh organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF020_chain_2_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]", "knowledge_formula_expr": "Formula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "parameters": {"OMN": 195.9, "SI": 0.65, "TI": 0.68, "WI": 0.279}, "problem": "In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Fresh organic matter carbon decomposition.\nUse the verified formula:\n\\[OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\\]\nIn calculation-friendly notation, this is:\nOMNC = WI * SI * TI * 0.027 * OMN.\nSubstitute the relevant values:\nWI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2.\nThe computed value is 0.652 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.652}.\n#### 0.652", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF016"} {"answer": "0.20", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF028", "CMF030"], "effective_parameters": {"B_above": 13156.0, "PDT": 39.3, "Y_grain": 4369.0}, "family": "growth_yield", "formula": {"expression": "(Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))", "id": "CMF042", "latex": "PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}", "name": "穗分配指数", "name_en": "Panicle partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF042_branch_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]", "knowledge_formula_expr": "Formula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "parameters": {"B_above": 13156.0, "PDT": 39.3, "Y_grain": 4369.0}, "problem": "For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Panicle partitioning index.\nUse the verified formula:\n\\[PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\\]\nIn calculation-friendly notation, this is:\nPIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39))).\nSubstitute the relevant values:\nY_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days.\nThe computed value is 0.20 fraction.\nTherefore, the final numeric answer is \\boxed{0.20}.\n#### 0.20", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF028"} {"answer": "40.80", "benchmark_version": "cropmath-v1", "change": {"new_value": 5430.0, "old_value": 4936.0, "parameter": "GrainYield"}, "distractor_formula_ids": ["CMF012", "CMF011"], "effective_parameters": {"GrainYield": 5430.0, "NUptake": 133.1}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "GrainYield / NUptake", "id": "CMF003", "latex": "NUE=\\frac{GrainYield}{NUptake}", "name": "氮利用效率", "name_en": "Nitrogen use efficiency", "precision": 2, "unit": "kg grain/kg N"}, "id": "CMF003_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]", "knowledge_formula_expr": "Formula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_name": "Relevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Ratio", "mode": "sensitivity", "parameters": {"GrainYield": 4936.0, "NUptake": 133.1}, "problem": "Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Nitrogen use efficiency.\nUse the verified formula:\n\\[NUE=\\frac{GrainYield}{NUptake}\\]\nIn calculation-friendly notation, this is:\nNUE = GrainYield / NUptake.\nSubstitute the relevant values:\nGrainYield=5430.0 kg/ha, NUptake=133.1 kg N/ha.\nThe computed value is 40.80 kg grain/kg N.\nTherefore, the final numeric answer is \\boxed{40.80}.\n#### 40.80", "split": "dev", "subject": "Nitrogen", "template_id": 1, "wrong_formula_id": "CMF012"} {"answer": "68.1", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF036", "CMF035"], "effective_parameters": {"SDEPTH": 3.3}, "family": "growth_yield", "formula": {"expression": "45 + 7 * SDEPTH", "id": "CMF028", "latex": "EMGDD=45+7SDEPTH", "name": "出苗热量需求", "name_en": "Emergence thermal requirement", "precision": 1, "unit": "C day"}, "id": "CMF028_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]", "knowledge_formula_expr": "Formula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear", "mode": "direct", "parameters": {"SDEPTH": 3.3}, "problem": "A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Emergence thermal requirement.\nUse the verified formula:\n\\[EMGDD=45+7SDEPTH\\]\nIn calculation-friendly notation, this is:\nEMGDD = 45 + 7 * SDEPTH.\nSubstitute the relevant values:\nSDEPTH=3.3 cm.\nThe computed value is 68.1 C day.\nTherefore, the final numeric answer is \\boxed{68.1}.\n#### 68.1", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF036"} {"answer": "0.000", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF055", "CMF058"], "effective_parameters": {"Tbase": 8.4, "Tmean": 3.3, "Topt": 22.7}, "family": "environmental_response", "formula": {"expression": "if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)", "id": "CMF060", "latex": "F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)", "name": "环境响应温度响应", "name_en": "environmental-response temperature response", "precision": 3, "unit": "fraction"}, "id": "CMF060_branch_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]", "knowledge_formula_expr": "Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise linear", "mode": "branch", "parameters": {"Tbase": 8.4, "Tmean": 3.3, "Topt": 22.7}, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response temperature response.\nUse the verified formula:\n\\[F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\\]\nIn calculation-friendly notation, this is:\nF_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0).\nSubstitute the relevant values:\nTmean=3.3 C, Tbase=8.4 C, Topt=22.7 C.\nThe computed value is 0.000 fraction.\nTherefore, the final numeric answer is \\boxed{0.000}.\n#### 0.000", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF055"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": {"new_value": 42.0, "old_value": 41.5, "parameter": "t_soil"}, "distractor_formula_ids": ["CMF018", "CMF016"], "effective_parameters": {"Q10": 3.41, "t_soil": 42.0}, "family": "methane_emission", "formula": {"expression": "Q10^((min(t_soil,30)-30)/10)", "id": "CMF023", "latex": "TI=Q_{10}^{(t_{sl}-30)/10}", "name": "甲烷温度指数", "name_en": "Methane temperature index", "precision": 3, "unit": "multiplier"}, "id": "CMF023_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]", "knowledge_formula_expr": "Formula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "parameters": {"Q10": 3.41, "t_soil": 41.5}, "problem": "In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane temperature index.\nUse the verified formula:\n\\[TI=Q_{10}^{(t_{sl}-30)/10}\\]\nIn calculation-friendly notation, this is:\nTI = Q10^((min(t_soil,30)-30)/10).\nSubstitute the relevant values:\nQ10=3.41 dimensionless, t_soil=42.0 C.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF018"} {"answer": "0.66", "benchmark_version": "cropmath-v1", "change": {"new_value": 8095.0, "old_value": 7039.0, "parameter": "Y_grain"}, "distractor_formula_ids": ["CMF028", "CMF047"], "effective_parameters": {"B_above": 12285.0, "Y_grain": 8095.0}, "family": "growth_yield", "formula": {"expression": "Y_grain / B_above", "id": "CMF034", "latex": "HI=\\frac{Y_{grain}}{B_{above}}", "name": "收获指数", "name_en": "Harvest index", "precision": 2, "unit": "fraction"}, "id": "CMF034_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]", "knowledge_formula_expr": "Formula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Harvest index (收获指数).\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Ratio", "mode": "sensitivity", "parameters": {"B_above": 12285.0, "Y_grain": 7039.0}, "problem": "Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Harvest index (收获指数).\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Harvest index.\nUse the verified formula:\n\\[HI=\\frac{Y_{grain}}{B_{above}}\\]\nIn calculation-friendly notation, this is:\nHI = Y_grain / B_above.\nSubstitute the relevant values:\nY_grain=8095.0 kg/ha, B_above=12285.0 kg/ha.\nThe computed value is 0.66 fraction.\nTherefore, the final numeric answer is \\boxed{0.66}.\n#### 0.66", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF028"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF062", "CMF060"], "effective_parameters": {"ExtremeT": 43.3, "MaxT": 34.5, "Tmax": 32.7}, "family": "environmental_response", "formula": {"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)", "id": "CMF058", "latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}", "name": "环境响应高温响应", "name_en": "environmental-response heat stress response", "precision": 3, "unit": "fraction"}, "id": "CMF058_branch_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]", "knowledge_formula_expr": "Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise linear stress", "mode": "branch", "parameters": {"ExtremeT": 43.3, "MaxT": 34.5, "Tmax": 32.7}, "problem": "Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response heat stress response.\nUse the verified formula:\n\\[F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\\]\nIn calculation-friendly notation, this is:\nF_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).\nSubstitute the relevant values:\nTmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF062"} {"answer": "0.648", "benchmark_version": "cropmath-v1", "change": {"new_value": 22.2, "old_value": 20.2, "parameter": "Tmean"}, "distractor_formula_ids": ["CMF059", "CMF056"], "effective_parameters": {"Tbase": 9.5, "Tmean": 22.2, "Topt": 29.1}, "family": "environmental_response", "formula": {"expression": "if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)", "id": "CMF060", "latex": "F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)", "name": "环境响应温度响应", "name_en": "environmental-response temperature response", "precision": 3, "unit": "fraction"}, "id": "CMF060_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]", "knowledge_formula_expr": "Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "parameters": {"Tbase": 9.5, "Tmean": 20.2, "Topt": 29.1}, "problem": "Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response temperature response.\nUse the verified formula:\n\\[F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\\]\nIn calculation-friendly notation, this is:\nF_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0).\nSubstitute the relevant values:\nTmean=22.2 C, Tbase=9.5 C, Topt=29.1 C.\nThe computed value is 0.648 fraction.\nTherefore, the final numeric answer is \\boxed{0.648}.\n#### 0.648", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF059"} {"answer": "0.14", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF033", "CMF046"], "effective_parameters": {"PDT": 30.5}, "family": "growth_yield", "formula": {"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)", "id": "CMF043", "latex": "PIRO=1-PISH", "name": "地下部分配指数", "name_en": "Root partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF043_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Residual from shoot partitioning", "mode": "chain_2", "parameters": {"PDT": 30.5}, "problem": "In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root partitioning index.\nUse the verified formula:\n\\[PIRO=1-PISH\\]\nIn calculation-friendly notation, this is:\nPIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).\nSubstitute the relevant values:\nPDT=30.5 physiological days.\nThe computed value is 0.14 fraction.\nTherefore, the final numeric answer is \\boxed{0.14}.\n#### 0.14", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF033"} {"answer": "0.157", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF002", "CMF011"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.102, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop water stress factor.\nUse the verified formula:\n\\[F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\\]\nIn calculation-friendly notation, this is:\nF_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Growth respiration.\nUse the verified formula:\n\\[RG=R_g\\cdot DTGA\\]\nIn calculation-friendly notation, this is:\nRG = Rg * DTGA.\nSubstitute the relevant values:\nRg=0.26 fraction, DTGA=148.5 kg/ha/day.\nThe computed value is 38.61 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{38.61}.\n#### 38.61", "split": "dev", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF052"} {"answer": "0.736", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF062", "CMF061"], "effective_parameters": {"I50A": 388.4, "I50B": 398.6, "TT": 1058.6, "Tsum": 1779.5, "fSolarMax": 0.89, "fSolar_water": 0.86}, "family": "environmental_response", "formula": {"expression": "min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)", "id": "CMF059", "latex": "fSolar=\\min(f_1,f_2)F_{solar,water}", "name": "环境响应光截获比例", "name_en": "environmental-response solar interception fraction", "precision": 3, "unit": "fraction"}, "id": "CMF059_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]", "knowledge_formula_expr": "Formula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response solar interception fraction (环境响应光截获比例).\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Double logistic", "mode": "direct", "parameters": {"I50A": 388.4, "I50B": 398.6, "TT": 1058.6, "Tsum": 1779.5, "fSolarMax": 0.89, "fSolar_water": 0.86}, "problem": "In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response solar interception fraction (环境响应光截获比例).\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response solar interception fraction.\nUse the verified formula:\n\\[fSolar=\\min(f_1,f_2)F_{solar,water}\\]\nIn calculation-friendly notation, this is:\nfSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1).\nSubstitute the relevant values:\nfSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction.\nThe computed value is 0.736 fraction.\nTherefore, the final numeric answer is \\boxed{0.736}.\n#### 0.736", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF062"} {"answer": "12.42", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.27, "old_value": 0.235, "parameter": "PS"}, "distractor_formula_ids": ["CMF027", "CMF051"], "effective_parameters": {"PS": 0.27, "Po": 10.5}, "family": "growth_yield", "formula": {"expression": "Po + sqrt(1 / PS)", "id": "CMF040", "latex": "P_c=P_o+\\sqrt{1/PS}", "name": "临界日长", "name_en": "Critical daylength", "precision": 2, "unit": "h"}, "id": "CMF040_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]", "knowledge_formula_expr": "Formula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_name": "Relevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Power response", "mode": "sensitivity", "parameters": {"PS": 0.235, "Po": 10.5}, "problem": "Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Critical daylength.\nUse the verified formula:\n\\[P_c=P_o+\\sqrt{1/PS}\\]\nIn calculation-friendly notation, this is:\nPc = Po + sqrt(1 / PS).\nSubstitute the relevant values:\nPo=10.5 h, PS=0.27 dimensionless.\nThe computed value is 12.42 h.\nTherefore, the final numeric answer is \\boxed{12.42}.\n#### 12.42", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF027"} {"answer": "101.31", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.23, "old_value": 0.2, "parameter": "Rg"}, "distractor_formula_ids": ["CMF050", "CMF039"], "effective_parameters": {"DTGA": 440.5, "Rg": 0.23}, "family": "growth_yield", "formula": {"expression": "Rg * DTGA", "id": "CMF046", "latex": "RG=R_g\\cdot DTGA", "name": "生长呼吸", "name_en": "Growth respiration", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF046_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]", "knowledge_formula_expr": "Formula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product", "mode": "sensitivity", "parameters": {"DTGA": 440.5, "Rg": 0.2}, "problem": "In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Growth respiration.\nUse the verified formula:\n\\[RG=R_g\\cdot DTGA\\]\nIn calculation-friendly notation, this is:\nRG = Rg * DTGA.\nSubstitute the relevant values:\nRg=0.23 fraction, DTGA=440.5 kg/ha/day.\nThe computed value is 101.31 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{101.31}.\n#### 101.31", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF050"} {"answer": "0.024", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF017", "CMF024"], "effective_parameters": {"P": 16.848, "Wr": 1725.2, "t_soil": 33.0}, "family": "methane_emission", "formula": {"expression": "min(0.7 * ln(t_soil) / Wr, 0.9) * P", "id": "CMF016", "latex": "E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P", "name": "甲烷冒泡排放", "name_en": "Methane ebullition emission", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF016_chain_2_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]", "knowledge_formula_expr": "Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "parameters": {"P": 16.848, "Wr": 1725.2, "t_soil": 33.0}, "problem": "Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane ebullition emission.\nUse the verified formula:\n\\[E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\\]\nIn calculation-friendly notation, this is:\nEbl = min(0.7 * ln(t_soil) / Wr, 0.9) * P.\nSubstitute the relevant values:\nP=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha.\nThe computed value is 0.024 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.024}.\n#### 0.024", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF017"} {"answer": "448.60", "benchmark_version": "cropmath-v1", "change": {"new_value": 2.74, "old_value": 2.49, "parameter": "LAI"}, "distractor_formula_ids": ["CMF011", "CMF004"], "effective_parameters": {"Coef_K": 0.74, "Coef_rou": 0.12, "LAI": 2.74, "Radi": 588.1}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)", "id": "CMF006", "latex": "PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}", "name": "冠层吸收光合有效辐射", "name_en": "carbon-nitrogen cycling canopy absorbed PAR", "precision": 2, "unit": "umol/m2/s"}, "id": "CMF006_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]", "knowledge_formula_expr": "Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "parameters": {"Coef_K": 0.74, "Coef_rou": 0.12, "LAI": 2.49, "Radi": 588.1}, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling canopy absorbed PAR.\nUse the verified formula:\n\\[PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\\]\nIn calculation-friendly notation, this is:\nPAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI).\nSubstitute the relevant values:\nRadi=588.1 W/m2, LAI=2.74 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless.\nThe computed value is 448.60 umol/m2/s.\nTherefore, the final numeric answer is \\boxed{448.60}.\n#### 448.60", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF011"} {"answer": "0.65", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF032", "CMF044"], "effective_parameters": {"B_above": 16042.0, "PDT": 35.1, "Y_grain": 5337.0}, "family": "growth_yield", "formula": {"expression": "1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))", "id": "CMF045", "latex": "PIS=1-PIGL-PIP", "name": "茎鞘分配指数", "name_en": "Stem partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF045_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Stem partitioning index.\nUse the verified formula:\n\\[PIS=1-PIGL-PIP\\]\nIn calculation-friendly notation, this is:\nPIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39))).\nSubstitute the relevant values:\nPDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha.\nThe computed value is 0.65 fraction.\nTherefore, the final numeric answer is \\boxed{0.65}.\n#### 0.65", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF032"} {"answer": "0.015", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF020", "CMF023"], "effective_parameters": {"P": 8.889, "Wr": 1500.3, "t_soil": 34.8}, "family": "methane_emission", "formula": {"expression": "min(0.7 * ln(t_soil) / Wr, 0.9) * P", "id": "CMF016", "latex": "E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P", "name": "甲烷冒泡排放", "name_en": "Methane ebullition emission", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF016_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]", "knowledge_formula_expr": "Formula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Logarithmic minimum", "mode": "direct", "parameters": {"P": 8.889, "Wr": 1500.3, "t_soil": 34.8}, "problem": "In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane ebullition emission.\nUse the verified formula:\n\\[E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\\]\nIn calculation-friendly notation, this is:\nEbl = min(0.7 * ln(t_soil) / Wr, 0.9) * P.\nSubstitute the relevant values:\nP=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha.\nThe computed value is 0.015 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.015}.\n#### 0.015", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF020"} {"answer": "6.134", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF022", "CMF024"], "effective_parameters": {"CH4RiceEfC": 0.244, "Ebl": 5.626, "P": 2.569, "W": 6214.2, "Wmax": 10951.2}, "family": "methane_emission", "formula": {"expression": "Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25", "id": "CMF017", "latex": "E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}", "name": "甲烷总排放", "name_en": "Total methane emission", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF017_chain_2_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]", "knowledge_formula_expr": "Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Chained sum", "mode": "chain_2", "parameters": {"CH4RiceEfC": 0.244, "Ebl": 5.626, "P": 2.569, "W": 6214.2, "Wmax": 10951.2}, "problem": "In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Total methane emission.\nUse the verified formula:\n\\[E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\\]\nIn calculation-friendly notation, this is:\nE = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25.\nSubstitute the relevant values:\nEbl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha.\nThe computed value is 6.134 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{6.134}.\n#### 6.134", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF022"} {"answer": "35.555", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF006", "CMF013"], "effective_parameters": {"Coef_alfa": 16.85, "Coef_beta": 158.8, "Leaf_N": 3.39, "PAR": 261.8}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "(Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)", "id": "CMF007", "latex": "P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}", "name": "最适光合速率", "name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "precision": 3, "unit": "umol CO2/m2/s"}, "id": "CMF007_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]", "knowledge_formula_expr": "Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "parameters": {"Coef_alfa": 16.85, "Coef_beta": 158.8, "Leaf_N": 3.39, "PAR": 261.8}, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling optimal photosynthetic rate.\nUse the verified formula:\n\\[P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\\]\nIn calculation-friendly notation, this is:\nPhoto_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta).\nSubstitute the relevant values:\nCoef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2.\nThe computed value is 35.555 umol CO2/m2/s.\nTherefore, the final numeric answer is \\boxed{35.555}.\n#### 35.555", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF006"} {"answer": "2.161", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF014", "CMF008"], "effective_parameters": {"Temp": 18.8, "is_soil_temp": 1.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)", "id": "CMF011", "latex": "F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T", "name": "土壤碳分解温度因子", "name_en": "carbon-nitrogen cycling soil carbon temperature factor", "precision": 3, "unit": "multiplier"}, "id": "CMF011_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]", "knowledge_formula_expr": "Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil carbon temperature factor.\nUse the verified formula:\n\\[F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\\]\nIn calculation-friendly notation, this is:\nF_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1).\nSubstitute the relevant values:\nTemp=18.8 C, is_soil_temp=1.0 0/1.\nThe computed value is 2.161 fraction.\nTherefore, the final numeric answer is \\boxed{2.161}.\n#### 2.161", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "wrong_formula_id": "CMF014"} {"answer": "0.698", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF010", "CMF009"], "effective_parameters": {"Temp": 5.9, "is_soil_temp": 1.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)", "id": "CMF011", "latex": "F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T", "name": "土壤碳分解温度因子", "name_en": "carbon-nitrogen cycling soil carbon temperature factor", "precision": 3, "unit": "multiplier"}, "id": "CMF011_branch_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]", "knowledge_formula_expr": "Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "parameters": {"Temp": 5.9, "is_soil_temp": 1.0}, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil carbon temperature factor.\nUse the verified formula:\n\\[F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\\]\nIn calculation-friendly notation, this is:\nF_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1).\nSubstitute the relevant values:\nTemp=5.9 C, is_soil_temp=1.0 0/1.\nThe computed value is 0.698 fraction.\nTherefore, the final numeric answer is \\boxed{0.698}.\n#### 0.698", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "wrong_formula_id": "CMF010"} {"answer": "0.814", "benchmark_version": "cropmath-v1", "change": {"new_value": 5.59, "old_value": 4.86, "parameter": "pH"}, "distractor_formula_ids": ["CMF014", "CMF003"], "effective_parameters": {"pH": 5.59}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "1 / (1 + exp(2.5 * (5 - pH)))", "id": "CMF010", "latex": "F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}", "name": "土壤pH分解因子", "name_en": "carbon-nitrogen cycling soil pH decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF010_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子).\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil pH decomposition factor.\nUse the verified formula:\n\\[F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\\]\nIn calculation-friendly notation, this is:\nF_pH = 1 / (1 + exp(2.5 * (5 - pH))).\nSubstitute the relevant values:\npH=5.59 pH.\nThe computed value is 0.814 fraction.\nTherefore, the final numeric answer is \\boxed{0.814}.\n#### 0.814", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "wrong_formula_id": "CMF014"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF037", "CMF053"], "effective_parameters": {"PDT": 15.7, "a": 0.023}, "family": "growth_yield", "formula": {"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))", "id": "CMF030", "latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)", "name": "生理年龄因子", "name_en": "Physiological age factor", "precision": 2, "unit": "fraction"}, "id": "CMF030_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]", "knowledge_formula_expr": "Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "parameters": {"PDT": 15.7, "a": 0.023}, "problem": "Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Physiological age factor.\nUse the verified formula:\n\\[F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\\]\nIn calculation-friendly notation, this is:\nFPA = if PDT<28: 1; else: exp(-a*(PDT-28)).\nSubstitute the relevant values:\nPDT=15.7 physiological days, a=0.023 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF037"} {"answer": "25.45", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF023", "CMF022"], "effective_parameters": {"Tair": 27.7}, "family": "methane_emission", "formula": {"expression": "4.4 + 0.76 * Tair", "id": "CMF024", "latex": "T_{soil}=4.4+0.76T_{air}", "name": "土壤温度估算", "name_en": "Soil temperature from air temperature", "precision": 2, "unit": "C"}, "id": "CMF024_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]", "knowledge_formula_expr": "Formula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_name": "Relevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear", "mode": "direct", "parameters": {"Tair": 27.7}, "problem": "In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil temperature from air temperature.\nUse the verified formula:\n\\[T_{soil}=4.4+0.76T_{air}\\]\nIn calculation-friendly notation, this is:\nTsoil = 4.4 + 0.76 * Tair.\nSubstitute the relevant values:\nTair=27.7 C.\nThe computed value is 25.45 C.\nTherefore, the final numeric answer is \\boxed{25.45}.\n#### 25.45", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF023"} {"answer": "0.374", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF036", "CMF050"], "effective_parameters": {"BFF": 0.69, "DTE": 0.331, "IE": 1.13, "PDT": 2.4, "RPE": 0.64}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "parameters": {"BFF": 0.69, "DTE": 0.331, "IE": 1.13, "PDT": 2.4, "RPE": 0.64}, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless.\nThe computed value is 0.374 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.374}.\n#### 0.374", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF036"} {"answer": "2.086", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF003", "CMF001"], "effective_parameters": {"Day_Length_hour": 11.13, "F_CO2": 1.21, "F_T": 0.42, "F_Water": 0.36, "Photo_Rate_opt": 23.71}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]", "knowledge_formula_expr": "Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "parameters": {"Day_Length_hour": 11.13, "F_CO2": 1.21, "F_T": 0.42, "F_Water": 0.36, "Photo_Rate_opt": 23.71}, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h.\nThe computed value is 2.086 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{2.086}.\n#### 2.086", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF003"} {"answer": "0.972", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF025", "CMF015"], "effective_parameters": {"OMS": 1220.1, "SI": 1.33, "TI": 0.435, "WI": 0.459}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.003 * OMS", "id": "CMF021", "latex": "OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS", "name": "土壤有机质碳分解量", "name_en": "Soil organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF021_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]", "knowledge_formula_expr": "Formula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Multiplicative", "mode": "direct", "parameters": {"OMS": 1220.1, "SI": 1.33, "TI": 0.435, "WI": 0.459}, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil organic matter carbon decomposition.\nUse the verified formula:\n\\[OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\\]\nIn calculation-friendly notation, this is:\nOMSC = WI * SI * TI * 0.003 * OMS.\nSubstitute the relevant values:\nWI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2.\nThe computed value is 0.972 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.972}.\n#### 0.972", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF025"} {"answer": "135.57", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF047", "CMF031"], "effective_parameters": {"GDD": 597.9, "SLAc": 155.4}, "family": "growth_yield", "formula": {"expression": "if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)", "id": "CMF053", "latex": "SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)", "name": "比叶面积", "name_en": "Specific leaf area", "precision": 2, "unit": "cm2/g"}, "id": "CMF053_branch_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]", "knowledge_formula_expr": "Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_name": "Relevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "parameters": {"GDD": 597.9, "SLAc": 155.4}, "problem": "For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Specific leaf area.\nUse the verified formula:\n\\[SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\\]\nIn calculation-friendly notation, this is:\nSLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200).\nSubstitute the relevant values:\nGDD=597.9 C day, SLAc=155.4 cm2/g.\nThe computed value is 135.57 cm2/g.\nTherefore, the final numeric answer is \\boxed{135.57}.\n#### 135.57", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF047"} {"answer": "1.519", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF023", "CMF015"], "effective_parameters": {"OMS": 1545.9, "SI": 0.92, "TI": 0.658, "WI": 0.541}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.003 * OMS", "id": "CMF021", "latex": "OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS", "name": "土壤有机质碳分解量", "name_en": "Soil organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF021_chain_2_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]", "knowledge_formula_expr": "Formula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "parameters": {"OMS": 1545.9, "SI": 0.92, "TI": 0.658, "WI": 0.541}, "problem": "A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil organic matter carbon decomposition.\nUse the verified formula:\n\\[OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\\]\nIn calculation-friendly notation, this is:\nOMSC = WI * SI * TI * 0.003 * OMS.\nSubstitute the relevant values:\nWI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2.\nThe computed value is 1.519 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{1.519}.\n#### 1.519", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF023"} {"answer": "28.104", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF014"], "effective_parameters": {"Day_Length_hour": 14.04, "F_CO2": 1.31, "F_T": 0.95, "F_Water": 0.69, "Photo_Rate_opt": 53.96}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h.\nThe computed value is 28.104 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{28.104}.\n#### 28.104", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF009"} {"answer": "3.42", "benchmark_version": "cropmath-v1", "change": {"new_value": 8153.0, "old_value": 7090.0, "parameter": "AWL"}, "distractor_formula_ids": ["CMF053", "CMF030"], "effective_parameters": {"AWL": 8153.0, "SLA": 0.00042}, "family": "growth_yield", "formula": {"expression": "SLA * AWL", "id": "CMF038", "latex": "LAI=SLA\\cdot AWL", "name": "比叶面积法叶面积指数", "name_en": "Leaf area index from SLA", "precision": 2, "unit": "m2/m2"}, "id": "CMF038_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]", "knowledge_formula_expr": "Formula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_name": "Relevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product", "mode": "sensitivity", "parameters": {"AWL": 7090.0, "SLA": 0.00042}, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Leaf area index from SLA.\nUse the verified formula:\n\\[LAI=SLA\\cdot AWL\\]\nIn calculation-friendly notation, this is:\nLAI = SLA * AWL.\nSubstitute the relevant values:\nSLA=0.00042 ha/kg, AWL=8153.0 kg/ha.\nThe computed value is 3.42 m2/m2.\nTherefore, the final numeric answer is \\boxed{3.42}.\n#### 3.42", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "wrong_formula_id": "CMF053"} {"answer": "11.71", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF044", "CMF028"], "effective_parameters": {"LAI": 2.82, "PAR": 14.18, "k": 0.62}, "family": "growth_yield", "formula": {"expression": "PAR * (1 - exp(-k * LAI))", "id": "CMF047", "latex": "IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})", "name": "截获光合有效辐射", "name_en": "Intercepted photosynthetically active radiation", "precision": 2, "unit": "MJ/m2/day"}, "id": "CMF047_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]", "knowledge_formula_expr": "Formula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day", "knowledge_name": "Relevant formula name: Intercepted photosynthetically active radiation (截获光合有效辐射).\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product with exponential saturation", "mode": "direct", "parameters": {"LAI": 2.82, "PAR": 14.18, "k": 0.62}, "problem": "A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Intercepted photosynthetically active radiation (截获光合有效辐射).\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Intercepted photosynthetically active radiation.\nUse the verified formula:\n\\[IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\\]\nIn calculation-friendly notation, this is:\nIPAR = PAR * (1 - exp(-k * LAI)).\nSubstitute the relevant values:\nPAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2.\nThe computed value is 11.71 MJ/m2/day.\nTherefore, the final numeric answer is \\boxed{11.71}.\n#### 11.71", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF044"} {"answer": "1.07", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.64, "old_value": 0.56, "parameter": "beta"}, "distractor_formula_ids": ["CMF046", "CMF026"], "effective_parameters": {"Cx": 377.0, "beta": 0.64}, "family": "growth_yield", "formula": {"expression": "1 + beta * ln(Cx / 340)", "id": "CMF029", "latex": "F_{CO2}=1+\\beta\\ln(C_x/340)", "name": "二氧化碳影响因子", "name_en": "CO2 response factor", "precision": 2, "unit": "multiplier"}, "id": "CMF029_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]", "knowledge_formula_expr": "Formula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "parameters": {"Cx": 377.0, "beta": 0.56}, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as CO2 response factor.\nUse the verified formula:\n\\[F_{CO2}=1+\\beta\\ln(C_x/340)\\]\nIn calculation-friendly notation, this is:\nFCO2 = 1 + beta * ln(Cx / 340).\nSubstitute the relevant values:\nCx=377.0 ppm, beta=0.64 dimensionless.\nThe computed value is 1.07 fraction.\nTherefore, the final numeric answer is \\boxed{1.07}.\n#### 1.07", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF046"} {"answer": "26.59", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF023", "CMF016"], "effective_parameters": {"Tair": 29.2}, "family": "methane_emission", "formula": {"expression": "4.4 + 0.76 * Tair", "id": "CMF024", "latex": "T_{soil}=4.4+0.76T_{air}", "name": "土壤温度估算", "name_en": "Soil temperature from air temperature", "precision": 2, "unit": "C"}, "id": "CMF024_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]", "knowledge_formula_expr": "Formula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_name": "Relevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear", "mode": "direct", "parameters": {"Tair": 29.2}, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil temperature from air temperature.\nUse the verified formula:\n\\[T_{soil}=4.4+0.76T_{air}\\]\nIn calculation-friendly notation, this is:\nTsoil = 4.4 + 0.76 * Tair.\nSubstitute the relevant values:\nTair=29.2 C.\nThe computed value is 26.59 C.\nTherefore, the final numeric answer is \\boxed{26.59}.\n#### 26.59", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF023"} {"answer": "0.173", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.41, "old_value": 0.45, "parameter": "RPE"}, "distractor_formula_ids": ["CMF044", "CMF029"], "effective_parameters": {"BFF": 1.07, "DTE": 0.422, "IE": 0.67, "PDT": 8.3, "RPE": 0.41}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "parameters": {"BFF": 1.07, "DTE": 0.422, "IE": 0.67, "PDT": 8.3, "RPE": 0.45}, "problem": "Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.41 fraction, BFF=1.07 dimensionless.\nThe computed value is 0.173 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.173}.\n#### 0.173", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF044"} {"answer": "1.454", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF014", "CMF005"], "effective_parameters": {"CO2": 613.0, "Coef_B_CO2": 0.77}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "1 + Coef_B_CO2 * ln(CO2 / 340)", "id": "CMF001", "latex": "F_{CO2}=1+B\\ln(CO2/340)", "name": "作物二氧化碳施肥因子", "name_en": "carbon-nitrogen cycling CO2 fertilization factor", "precision": 3, "unit": "multiplier"}, "id": "CMF001_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling CO2 fertilization factor.\nUse the verified formula:\n\\[F_{CO2}=1+B\\ln(CO2/340)\\]\nIn calculation-friendly notation, this is:\nF_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340).\nSubstitute the relevant values:\nCO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless.\nThe computed value is 1.454 fraction.\nTherefore, the final numeric answer is \\boxed{1.454}.\n#### 1.454", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF014"} {"answer": "17.20", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF027", "CMF043"], "effective_parameters": {"Am": 28.0, "FCO2": 1.16, "FPA": 0.85, "FT": 0.89, "NNI": 0.84, "WDF": 0.7}, "family": "growth_yield", "formula": {"expression": "Am * FCO2 * FPA * FT * min(NNI, WDF)", "id": "CMF026", "latex": "AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)", "name": "实际最大同化速率", "name_en": "Actual maximum assimilation rate", "precision": 2, "unit": "kg CO2/ha/h"}, "id": "CMF026_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_name": "Relevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "parameters": {"Am": 28.0, "FCO2": 1.16, "FPA": 0.85, "FT": 0.89, "NNI": 0.84, "WDF": 0.7}, "problem": "In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual maximum assimilation rate.\nUse the verified formula:\n\\[AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nAMAX = Am * FCO2 * FPA * FT * min(NNI, WDF).\nSubstitute the relevant values:\nAm=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction.\nThe computed value is 17.20 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{17.20}.\n#### 17.20", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF027"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF028", "CMF039"], "effective_parameters": {"P": 10.46, "PS": 0.095, "Po": 10.5}, "family": "growth_yield", "formula": {"expression": "if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0", "id": "CMF050", "latex": "RPE=1,\\;1-PS(P-P_o)^2,\\;0", "name": "相对光周期效应", "name_en": "Relative photoperiod effectiveness", "precision": 2, "unit": "fraction"}, "id": "CMF050_branch_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]", "knowledge_formula_expr": "Formula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Relative photoperiod effectiveness (相对光周期效应).\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise quadratic", "mode": "branch", "parameters": {"P": 10.46, "PS": 0.095, "Po": 10.5}, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative photoperiod effectiveness (相对光周期效应).\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Relative photoperiod effectiveness.\nUse the verified formula:\n\\[RPE=1,\\;1-PS(P-P_o)^2,\\;0\\]\nIn calculation-friendly notation, this is:\nRPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0.\nSubstitute the relevant values:\nP=10.46 h, Po=10.5 h, PS=0.095 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF028"} {"answer": "0.182", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF023", "CMF025"], "effective_parameters": {"CH4RiceEfC": 0.217, "W": 6048.7, "Wmax": 11870.1}, "family": "methane_emission", "formula": {"expression": "CH4RiceEfC * (1 - W / Wmax) ** 0.25", "id": "CMF019", "latex": "F_p=C(1-W/W_{max})^{0.25}", "name": "植株传输因子", "name_en": "Plant methane transport factor", "precision": 3, "unit": "fraction"}, "id": "CMF019_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]", "knowledge_formula_expr": "Formula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Plant methane transport factor (植株传输因子).\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Power attenuation", "mode": "direct", "parameters": {"CH4RiceEfC": 0.217, "W": 6048.7, "Wmax": 11870.1}, "problem": "In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Plant methane transport factor (植株传输因子).\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Plant methane transport factor.\nUse the verified formula:\n\\[F_p=C(1-W/W_{max})^{0.25}\\]\nIn calculation-friendly notation, this is:\nFp = CH4RiceEfC * (1 - W / Wmax) ** 0.25.\nSubstitute the relevant values:\nCH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha.\nThe computed value is 0.182 fraction.\nTherefore, the final numeric answer is \\boxed{0.182}.\n#### 0.182", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF023"} {"answer": "0.0022", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF041", "CMF037"], "effective_parameters": {"NNI": 0.58, "RP": 0.0038, "WDF": 0.73}, "family": "growth_yield", "formula": {"expression": "RP * min(NNI, WDF)", "id": "CMF052", "latex": "R=RP\\cdot\\min(NNI,WDF)", "name": "实际叶面积相对增长率", "name_en": "Actual LAI relative growth rate", "precision": 4, "unit": "d-1"}, "id": "CMF052_chain_2_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_name": "Relevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "parameters": {"NNI": 0.58, "RP": 0.0038, "WDF": 0.73}, "problem": "Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual LAI relative growth rate.\nUse the verified formula:\n\\[R=RP\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nR = RP * min(NNI, WDF).\nSubstitute the relevant values:\nRP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction.\nThe computed value is 0.0022 d-1.\nTherefore, the final numeric answer is \\boxed{0.0022}.\n#### 0.0022", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF041"} {"answer": "24.14", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF049", "CMF037"], "effective_parameters": {"ABIOMASS": 5647.0, "PDT": 55.4, "Q10": 1.88, "T": 32.4, "To": 29.5}, "family": "growth_yield", "formula": {"expression": "(0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)", "id": "CMF048", "latex": "RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}", "name": "维持呼吸", "name_en": "Maintenance respiration", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF048_chain_2_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]", "knowledge_formula_expr": "Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Product with Q10 temperature response", "mode": "chain_2", "parameters": {"ABIOMASS": 5647.0, "PDT": 55.4, "Q10": 1.88, "T": 32.4, "To": 29.5}, "problem": "Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration.\nUse the verified formula:\n\\[RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\\]\nIn calculation-friendly notation, this is:\nRM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10).\nSubstitute the relevant values:\nPDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C.\nThe computed value is 24.14 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{24.14}.\n#### 24.14", "split": "dev", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF049"} {"answer": "243.95", "benchmark_version": "cropmath-v1", "change": {"new_value": 187.7, "old_value": 170.6, "parameter": "SLAc"}, "distractor_formula_ids": ["CMF031", "CMF027"], "effective_parameters": {"GDD": 488.0, "SLAc": 187.7}, "family": "growth_yield", "formula": {"expression": "if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)", "id": "CMF053", "latex": "SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)", "name": "比叶面积", "name_en": "Specific leaf area", "precision": 2, "unit": "cm2/g"}, "id": "CMF053_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]", "knowledge_formula_expr": "Formula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_name": "Relevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "parameters": {"GDD": 488.0, "SLAc": 170.6}, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Specific leaf area.\nUse the verified formula:\n\\[SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\\]\nIn calculation-friendly notation, this is:\nSLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200).\nSubstitute the relevant values:\nGDD=488.0 C day, SLAc=187.7 cm2/g.\nThe computed value is 243.95 cm2/g.\nTherefore, the final numeric answer is \\boxed{243.95}.\n#### 243.95", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "wrong_formula_id": "CMF031"} {"answer": "0.025", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.53, "old_value": 0.48, "parameter": "S_Water"}, "distractor_formula_ids": ["CMF057", "CMF059"], "effective_parameters": {"ARID": 1.84, "S_Water": 0.53}, "family": "environmental_response", "formula": {"expression": "max(0, 1 - S_Water * ARID)", "id": "CMF061", "latex": "F_W=\\max(0,1-S_WARID)", "name": "环境响应水分响应", "name_en": "environmental-response water stress response", "precision": 3, "unit": "response factor"}, "id": "CMF061_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]", "knowledge_formula_expr": "Formula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response water stress response (环境响应水分响应).\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "parameters": {"ARID": 1.84, "S_Water": 0.48}, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response water stress response (环境响应水分响应).\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response water stress response.\nUse the verified formula:\n\\[F_W=\\max(0,1-S_WARID)\\]\nIn calculation-friendly notation, this is:\nF_Water = max(0, 1 - S_Water * ARID).\nSubstitute the relevant values:\nARID=1.84 dimensionless, S_Water=0.53 dimensionless.\nThe computed value is 0.025 fraction.\nTherefore, the final numeric answer is \\boxed{0.025}.\n#### 0.025", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF057"} {"answer": "0.936", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF024", "CMF020"], "effective_parameters": {"WaterC": 0.455}, "family": "methane_emission", "formula": {"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)", "id": "CMF025", "latex": "WI=0.49e^{3.88W_c-5.4W_c^2}", "name": "水分分解因子", "name_en": "Water-content decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF025_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]", "knowledge_formula_expr": "Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "parameters": {"WaterC": 0.455}, "problem": "Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Water-content decomposition factor.\nUse the verified formula:\n\\[WI=0.49e^{3.88W_c-5.4W_c^2}\\]\nIn calculation-friendly notation, this is:\nWI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).\nSubstitute the relevant values:\nWaterC=0.455 fraction.\nThe computed value is 0.936 fraction.\nTherefore, the final numeric answer is \\boxed{0.936}.\n#### 0.936", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF024"} {"answer": "0.094", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF013", "CMF002"], "effective_parameters": {"Coef_Plant_TN_uptake": 0.204, "DVI": 1.28, "NPP": 8.58, "N_requirement_max_day": 2.99, "N_shoot": 5.0, "Ratio_N_C_photo_product": 0.011}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)", "id": "CMF004", "latex": "N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)", "name": "作物需氮量", "name_en": "carbon-nitrogen cycling crop nitrogen demand", "precision": 3, "unit": "g N/m2/day"}, "id": "CMF004_branch_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]", "knowledge_formula_expr": "Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise minimum", "mode": "branch", "parameters": {"Coef_Plant_TN_uptake": 0.204, "DVI": 1.28, "NPP": 8.58, "N_requirement_max_day": 2.99, "N_shoot": 5.0, "Ratio_N_C_photo_product": 0.011}, "problem": "For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop nitrogen demand.\nUse the verified formula:\n\\[N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\\]\nIn calculation-friendly notation, this is:\nN_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP).\nSubstitute the relevant values:\nN_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless.\nThe computed value is 0.094 g N/m2/day.\nTherefore, the final numeric answer is \\boxed{0.094}.\n#### 0.094", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "wrong_formula_id": "CMF013"} {"answer": "1.286", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF062", "CMF055"], "effective_parameters": {"CO2": 557.0, "CO2_RUE": 0.138}, "family": "environmental_response", "formula": {"expression": "if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)", "id": "CMF057", "latex": "F_{CO2}=f(CO2,CO2_{RUE})", "name": "环境响应二氧化碳响应", "name_en": "environmental-response CO2 response", "precision": 3, "unit": "multiplier"}, "id": "CMF057_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]", "knowledge_formula_expr": "Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "parameters": {"CO2": 557.0, "CO2_RUE": 0.138}, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response CO2 response.\nUse the verified formula:\n\\[F_{CO2}=f(CO2,CO2_{RUE})\\]\nIn calculation-friendly notation, this is:\nF_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1).\nSubstitute the relevant values:\nCO2=557.0 ppm, CO2_RUE=0.138 fraction.\nThe computed value is 1.286 fraction.\nTherefore, the final numeric answer is \\boxed{1.286}.\n#### 1.286", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF062"} {"answer": "0.0082", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF035", "CMF030"], "effective_parameters": {"PDT": 9.3}, "family": "growth_yield", "formula": {"expression": "0.0091 - 0.0001 * PDT", "id": "CMF049", "latex": "RM(T_o)=0.0091-0.0001PDT", "name": "最适温度维持呼吸系数", "name_en": "Maintenance respiration coefficient at optimum temperature", "precision": 4, "unit": "d-1"}, "id": "CMF049_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_name": "Relevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear decline", "mode": "direct", "parameters": {"PDT": 9.3}, "problem": "Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration coefficient at optimum temperature.\nUse the verified formula:\n\\[RM(T_o)=0.0091-0.0001PDT\\]\nIn calculation-friendly notation, this is:\nRMTo = 0.0091 - 0.0001 * PDT.\nSubstitute the relevant values:\nPDT=9.3 physiological days.\nThe computed value is 0.0082 d-1.\nTherefore, the final numeric answer is \\boxed{0.0082}.\n#### 0.0082", "split": "dev", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF035"} {"answer": "1.02", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF044", "CMF047"], "effective_parameters": {"AWL": 2747.0, "SLA": 0.00037}, "family": "growth_yield", "formula": {"expression": "SLA * AWL", "id": "CMF038", "latex": "LAI=SLA\\cdot AWL", "name": "比叶面积法叶面积指数", "name_en": "Leaf area index from SLA", "precision": 2, "unit": "m2/m2"}, "id": "CMF038_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]", "knowledge_formula_expr": "Formula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_name": "Relevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"AWL": 2747.0, "SLA": 0.00037}, "problem": "Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Leaf area index from SLA.\nUse the verified formula:\n\\[LAI=SLA\\cdot AWL\\]\nIn calculation-friendly notation, this is:\nLAI = SLA * AWL.\nSubstitute the relevant values:\nSLA=0.00037 ha/kg, AWL=2747.0 kg/ha.\nThe computed value is 1.02 m2/m2.\nTherefore, the final numeric answer is \\boxed{1.02}.\n#### 1.02", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF044"} {"answer": "0.592", "benchmark_version": "cropmath-v1", "change": {"new_value": -103.8, "old_value": -90.3, "parameter": "Eh"}, "distractor_formula_ids": ["CMF016", "CMF024"], "effective_parameters": {"Eh": -103.8}, "family": "methane_emission", "formula": {"expression": "if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))", "id": "CMF018", "latex": "F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}", "name": "氧化还原电位因子", "name_en": "Redox potential factor", "precision": 3, "unit": "fraction"}, "id": "CMF018_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]", "knowledge_formula_expr": "Formula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "parameters": {"Eh": -90.3}, "problem": "Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Redox potential factor.\nUse the verified formula:\n\\[F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\\]\nIn calculation-friendly notation, this is:\nFEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150)).\nSubstitute the relevant values:\nEh=-103.8 mV.\nThe computed value is 0.592 fraction.\nTherefore, the final numeric answer is \\boxed{0.592}.\n#### 0.592", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF016"} {"answer": "2.409", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF024", "CMF016"], "effective_parameters": {"Com": 6.896, "Cr": 63.36, "FEh": 0.354, "TI": 0.289}, "family": "methane_emission", "formula": {"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))", "id": "CMF022", "latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))", "name": "甲烷产生量", "name_en": "Methane production", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF022_chain_2_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]", "knowledge_formula_expr": "Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "parameters": {"Com": 6.896, "Cr": 63.36, "FEh": 0.354, "TI": 0.289}, "problem": "In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane production.\nUse the verified formula:\n\\[P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\\]\nIn calculation-friendly notation, this is:\nP = max(0, 0.27 * FEh * (TI * Cr + Com)).\nSubstitute the relevant values:\nFEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day.\nThe computed value is 2.409 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{2.409}.\n#### 2.409", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF024"} {"answer": "0.58", "benchmark_version": "cropmath-v1", "change": {"new_value": 9000.0, "old_value": 9991.0, "parameter": "B_above"}, "distractor_formula_ids": ["CMF043", "CMF026"], "effective_parameters": {"B_above": 9000.0, "Y_grain": 5253.0}, "family": "growth_yield", "formula": {"expression": "Y_grain / B_above", "id": "CMF034", "latex": "HI=\\frac{Y_{grain}}{B_{above}}", "name": "收获指数", "name_en": "Harvest index", "precision": 2, "unit": "fraction"}, "id": "CMF034_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]", "knowledge_formula_expr": "Formula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Harvest index (收获指数).\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Ratio", "mode": "sensitivity", "parameters": {"B_above": 9991.0, "Y_grain": 5253.0}, "problem": "In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Harvest index (收获指数).\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Harvest index.\nUse the verified formula:\n\\[HI=\\frac{Y_{grain}}{B_{above}}\\]\nIn calculation-friendly notation, this is:\nHI = Y_grain / B_above.\nSubstitute the relevant values:\nY_grain=5253.0 kg/ha, B_above=9000.0 kg/ha.\nThe computed value is 0.58 fraction.\nTherefore, the final numeric answer is \\boxed{0.58}.\n#### 0.58", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF043"} {"answer": "0.961", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.4, "old_value": 1.27, "parameter": "BD"}, "distractor_formula_ids": ["CMF005", "CMF003"], "effective_parameters": {"BD": 1.4, "Saturate_Water": 0.48, "Soil_Water": 0.411}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)", "id": "CMF012", "latex": "F_W=0.49e^{3.88W-5.4W^2}", "name": "土壤碳分解水分因子", "name_en": "carbon-nitrogen cycling soil carbon water factor", "precision": 3, "unit": "fraction"}, "id": "CMF012_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]", "knowledge_formula_expr": "Formula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子).\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "parameters": {"BD": 1.27, "Saturate_Water": 0.48, "Soil_Water": 0.411}, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子).\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil carbon water factor.\nUse the verified formula:\n\\[F_W=0.49e^{3.88W-5.4W^2}\\]\nIn calculation-friendly notation, this is:\nF_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2).\nSubstitute the relevant values:\nSoil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.4 g/cm3.\nThe computed value is 0.961 fraction.\nTherefore, the final numeric answer is \\boxed{0.961}.\n#### 0.961", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "wrong_formula_id": "CMF005"} {"answer": "18.30", "benchmark_version": "cropmath-v1", "change": {"new_value": 30.0, "old_value": 33.3, "parameter": "Tmean"}, "distractor_formula_ids": ["CMF062", "CMF058"], "effective_parameters": {"Tbase": 11.7, "Tmean": 30.0}, "family": "environmental_response", "formula": {"expression": "max(Tmean - Tbase, 0)", "id": "CMF056", "latex": "dTT=\\max(T_{mean}-T_{base},0)", "name": "每日有效积温", "name_en": "environmental-response daily thermal time", "precision": 2, "unit": "C day"}, "id": "CMF056_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]", "knowledge_formula_expr": "Formula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "parameters": {"Tbase": 11.7, "Tmean": 33.3}, "problem": "Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily thermal time.\nUse the verified formula:\n\\[dTT=\\max(T_{mean}-T_{base},0)\\]\nIn calculation-friendly notation, this is:\ndTT = max(Tmean - Tbase, 0).\nSubstitute the relevant values:\nTmean=30.0 C, Tbase=11.7 C.\nThe computed value is 18.30 C day.\nTherefore, the final numeric answer is \\boxed{18.30}.\n#### 18.30", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF062"} {"answer": "487.96", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF001", "CMF008"], "effective_parameters": {"Coef_K": 0.38, "Coef_rou": 0.23, "LAI": 2.51, "Radi": 485.8}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)", "id": "CMF006", "latex": "PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}", "name": "冠层吸收光合有效辐射", "name_en": "carbon-nitrogen cycling canopy absorbed PAR", "precision": 2, "unit": "umol/m2/s"}, "id": "CMF006_chain_2_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]", "knowledge_formula_expr": "Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "parameters": {"Coef_K": 0.38, "Coef_rou": 0.23, "LAI": 2.51, "Radi": 485.8}, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling canopy absorbed PAR.\nUse the verified formula:\n\\[PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\\]\nIn calculation-friendly notation, this is:\nPAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI).\nSubstitute the relevant values:\nRadi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless.\nThe computed value is 487.96 umol/m2/s.\nTherefore, the final numeric answer is \\boxed{487.96}.\n#### 487.96", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF001"} {"answer": "0.094", "benchmark_version": "cropmath-v1", "change": {"new_value": 42.0, "old_value": 43.8, "parameter": "ExtremeT"}, "distractor_formula_ids": ["CMF056", "CMF062"], "effective_parameters": {"ExtremeT": 42.0, "MaxT": 33.5, "Tmax": 41.2}, "family": "environmental_response", "formula": {"expression": "if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)", "id": "CMF058", "latex": "F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}", "name": "环境响应高温响应", "name_en": "environmental-response heat stress response", "precision": 3, "unit": "fraction"}, "id": "CMF058_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]", "knowledge_formula_expr": "Formula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "parameters": {"ExtremeT": 43.8, "MaxT": 33.5, "Tmax": 41.2}, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response heat stress response.\nUse the verified formula:\n\\[F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\\]\nIn calculation-friendly notation, this is:\nF_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0).\nSubstitute the relevant values:\nTmax=41.2 C, MaxT=33.5 C, ExtremeT=42.0 C.\nThe computed value is 0.094 fraction.\nTherefore, the final numeric answer is \\boxed{0.094}.\n#### 0.094", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF056"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 22.9, "old_value": 19.9, "parameter": "Tmean"}, "distractor_formula_ids": ["CMF054", "CMF047"], "effective_parameters": {"PDT": 37.4, "Qt": 17.4, "Tmean": 22.9}, "family": "growth_yield", "formula": {"expression": "if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1", "id": "CMF039", "latex": "LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}", "name": "低温结实因子", "name_en": "Low-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF039_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]", "knowledge_formula_expr": "Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "parameters": {"PDT": 37.4, "Qt": 17.4, "Tmean": 19.9}, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Low-temperature fertility factor.\nUse the verified formula:\n\\[LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\\]\nIn calculation-friendly notation, this is:\nLTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1.\nSubstitute the relevant values:\nPDT=37.4 physiological days, Tmean=22.9 C, Qt=17.4 C day.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Temperature stress", "template_id": 0, "wrong_formula_id": "CMF054"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 23.0, "old_value": 22.0, "parameter": "Tmean"}, "distractor_formula_ids": ["CMF026", "CMF047"], "effective_parameters": {"PDT": 33.6, "Qt": 3.6, "Tmean": 23.0}, "family": "growth_yield", "formula": {"expression": "if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1", "id": "CMF039", "latex": "LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}", "name": "低温结实因子", "name_en": "Low-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF039_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]", "knowledge_formula_expr": "Formula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "parameters": {"PDT": 33.6, "Qt": 3.6, "Tmean": 22.0}, "problem": "In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Low-temperature fertility factor.\nUse the verified formula:\n\\[LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\\]\nIn calculation-friendly notation, this is:\nLTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1.\nSubstitute the relevant values:\nPDT=33.6 physiological days, Tmean=23.0 C, Qt=3.6 C day.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Temperature stress", "template_id": 2, "wrong_formula_id": "CMF026"} {"answer": "68.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF037", "CMF035"], "effective_parameters": {"ABIOMASS": 8842.0, "PDT": 20.6, "Q10": 1.88, "T": 30.6, "To": 29.2}, "family": "growth_yield", "formula": {"expression": "(0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)", "id": "CMF048", "latex": "RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}", "name": "维持呼吸", "name_en": "Maintenance respiration", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF048_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]", "knowledge_formula_expr": "Formula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "parameters": {"ABIOMASS": 8842.0, "PDT": 20.6, "Q10": 1.88, "T": 30.6, "To": 29.2}, "problem": "In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration.\nUse the verified formula:\n\\[RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\\]\nIn calculation-friendly notation, this is:\nRM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10).\nSubstitute the relevant values:\nPDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C.\nThe computed value is 68.00 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{68.00}.\n#### 68.00", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF037"} {"answer": "0.576", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF038", "CMF036"], "effective_parameters": {"BFF": 0.67, "DTE": 0.86, "IE": 1.09, "PDT": 33.7, "RPE": 0.59}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_chain_2_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "parameters": {"BFF": 0.67, "DTE": 0.86, "IE": 1.09, "PDT": 33.7, "RPE": 0.59}, "problem": "Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless.\nThe computed value is 0.576 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.576}.\n#### 0.576", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF038"} {"answer": "1.240", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF005"], "effective_parameters": {"CO2": 550.0, "Coef_B_CO2": 0.5}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "1 + Coef_B_CO2 * ln(CO2 / 340)", "id": "CMF001", "latex": "F_{CO2}=1+B\\ln(CO2/340)", "name": "作物二氧化碳施肥因子", "name_en": "carbon-nitrogen cycling CO2 fertilization factor", "precision": 3, "unit": "multiplier"}, "id": "CMF001_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]", "knowledge_formula_expr": "Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Logarithmic", "mode": "direct", "parameters": {"CO2": 550.0, "Coef_B_CO2": 0.5}, "problem": "Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling CO2 fertilization factor.\nUse the verified formula:\n\\[F_{CO2}=1+B\\ln(CO2/340)\\]\nIn calculation-friendly notation, this is:\nF_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340).\nSubstitute the relevant values:\nCO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless.\nThe computed value is 1.240 fraction.\nTherefore, the final numeric answer is \\boxed{1.240}.\n#### 1.240", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "wrong_formula_id": "CMF009"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF010"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.339, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop water stress factor.\nUse the verified formula:\n\\[F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\\]\nIn calculation-friendly notation, this is:\nF_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h.\nThe computed value is 7.802 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{7.802}.\n#### 7.802", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF005"} {"answer": "0.678", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF013", "CMF014"], "effective_parameters": {"Biom_plant_yesterday": 691.1, "Coef_Rm": 0.0013, "DVI": 1.63, "Tmn": 23.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "parameters": {"Biom_plant_yesterday": 691.1, "Coef_Rm": 0.0013, "DVI": 1.63, "Tmn": 23.0}, "problem": "Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless.\nThe computed value is 0.678 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.678}.\n#### 0.678", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "wrong_formula_id": "CMF013"} {"answer": "4.020", "benchmark_version": "cropmath-v1", "change": {"new_value": 39.5, "old_value": 35.91, "parameter": "Photo_Rate_opt"}, "distractor_formula_ids": ["CMF003", "CMF006"], "effective_parameters": {"Day_Length_hour": 12.71, "F_CO2": 0.97, "F_T": 0.49, "F_Water": 0.39, "Photo_Rate_opt": 39.5}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]", "knowledge_formula_expr": "Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "parameters": {"Day_Length_hour": 12.71, "F_CO2": 0.97, "F_T": 0.49, "F_Water": 0.39, "Photo_Rate_opt": 35.91}, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=39.5 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h.\nThe computed value is 4.020 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{4.020}.\n#### 4.020", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF003"} {"answer": "6.640", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF013", "CMF006"], "effective_parameters": {"Day_Length_hour": 13.35, "F_CO2": 1.23, "F_T": 0.54, "F_Water": 0.54, "Photo_Rate_opt": 32.1}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_chain_2_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]", "knowledge_formula_expr": "Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "parameters": {"Day_Length_hour": 13.35, "F_CO2": 1.23, "F_T": 0.54, "F_Water": 0.54, "Photo_Rate_opt": 32.1}, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h.\nThe computed value is 6.640 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{6.640}.\n#### 6.640", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF013"} {"answer": "0.028", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.8, "old_value": 1.65, "parameter": "DVI"}, "distractor_formula_ids": ["CMF012", "CMF013"], "effective_parameters": {"Biom_plant_yesterday": 131.0, "Coef_Rm": 0.0005, "DVI": 1.8, "Tmn": 17.4}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)", "id": "CMF008", "latex": "R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}", "name": "植株维持呼吸", "name_en": "carbon-nitrogen cycling plant maintenance respiration", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF008_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]", "knowledge_formula_expr": "Formula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "parameters": {"Biom_plant_yesterday": 131.0, "Coef_Rm": 0.0005, "DVI": 1.65, "Tmn": 17.4}, "problem": "In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling plant maintenance respiration.\nUse the verified formula:\n\\[R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\\]\nIn calculation-friendly notation, this is:\nRm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1).\nSubstitute the relevant values:\nBiom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.8 dimensionless.\nThe computed value is 0.028 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.028}.\n#### 0.028", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "wrong_formula_id": "CMF012"} {"answer": "10.44", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF029", "CMF035"], "effective_parameters": {"Am": 53.7, "FCO2": 1.27, "FPA": 0.52, "FT": 0.64, "NNI": 0.46, "WDF": 0.89}, "family": "growth_yield", "formula": {"expression": "Am * FCO2 * FPA * FT * min(NNI, WDF)", "id": "CMF026", "latex": "AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)", "name": "实际最大同化速率", "name_en": "Actual maximum assimilation rate", "precision": 2, "unit": "kg CO2/ha/h"}, "id": "CMF026_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]", "knowledge_formula_expr": "Formula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_name": "Relevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "parameters": {"Am": 53.7, "FCO2": 1.27, "FPA": 0.52, "FT": 0.64, "NNI": 0.46, "WDF": 0.89}, "problem": "Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual maximum assimilation rate.\nUse the verified formula:\n\\[AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nAMAX = Am * FCO2 * FPA * FT * min(NNI, WDF).\nSubstitute the relevant values:\nAm=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction.\nThe computed value is 10.44 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{10.44}.\n#### 10.44", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF029"} {"answer": "18.10", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF061", "CMF062"], "effective_parameters": {"Tbase": 11.5, "Tmean": 29.6}, "family": "environmental_response", "formula": {"expression": "max(Tmean - Tbase, 0)", "id": "CMF056", "latex": "dTT=\\max(T_{mean}-T_{base},0)", "name": "每日有效积温", "name_en": "environmental-response daily thermal time", "precision": 2, "unit": "C day"}, "id": "CMF056_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]", "knowledge_formula_expr": "Formula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Maximum linear", "mode": "direct", "parameters": {"Tbase": 11.5, "Tmean": 29.6}, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily thermal time.\nUse the verified formula:\n\\[dTT=\\max(T_{mean}-T_{base},0)\\]\nIn calculation-friendly notation, this is:\ndTT = max(Tmean - Tbase, 0).\nSubstitute the relevant values:\nTmean=29.6 C, Tbase=11.5 C.\nThe computed value is 18.10 C day.\nTherefore, the final numeric answer is \\boxed{18.10}.\n#### 18.10", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF061"} {"answer": "8.46", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.57, "old_value": 0.52, "parameter": "k"}, "distractor_formula_ids": ["CMF034", "CMF043"], "effective_parameters": {"LAI": 1.98, "PAR": 12.51, "k": 0.57}, "family": "growth_yield", "formula": {"expression": "PAR * (1 - exp(-k * LAI))", "id": "CMF047", "latex": "IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})", "name": "截获光合有效辐射", "name_en": "Intercepted photosynthetically active radiation", "precision": 2, "unit": "MJ/m2/day"}, "id": "CMF047_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]", "knowledge_formula_expr": "Formula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day", "knowledge_name": "Relevant formula name: Intercepted photosynthetically active radiation (截获光合有效辐射).\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product with exponential saturation", "mode": "sensitivity", "parameters": {"LAI": 1.98, "PAR": 12.51, "k": 0.52}, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Intercepted photosynthetically active radiation (截获光合有效辐射).\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Intercepted photosynthetically active radiation.\nUse the verified formula:\n\\[IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\\]\nIn calculation-friendly notation, this is:\nIPAR = PAR * (1 - exp(-k * LAI)).\nSubstitute the relevant values:\nPAR=12.51 MJ/m2/day, k=0.57 dimensionless, LAI=1.98 m2/m2.\nThe computed value is 8.46 MJ/m2/day.\nTherefore, the final numeric answer is \\boxed{8.46}.\n#### 8.46", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF034"} {"answer": "20.32", "benchmark_version": "cropmath-v1", "change": {"new_value": 120.0, "old_value": 113.3, "parameter": "RM"}, "distractor_formula_ids": ["CMF031", "CMF050"], "effective_parameters": {"DTGA": 194.1, "RG": 54.8, "RM": 120.0, "b": 0.05}, "family": "growth_yield", "formula": {"expression": "(DTGA - RM - RG) / (1 - b)", "id": "CMF033", "latex": "GCR=\\frac{DTGA-RM-RG}{1-b}", "name": "日干物质增量", "name_en": "Daily dry matter increment", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF033_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]", "knowledge_formula_expr": "Formula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Subtraction and conversion", "mode": "sensitivity", "parameters": {"DTGA": 194.1, "RG": 54.8, "RM": 113.3, "b": 0.05}, "problem": "In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily dry matter increment.\nUse the verified formula:\n\\[GCR=\\frac{DTGA-RM-RG}{1-b}\\]\nIn calculation-friendly notation, this is:\nGCR = (DTGA - RM - RG) / (1 - b).\nSubstitute the relevant values:\nDTGA=194.1 kg/ha/day, RM=120.0 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction.\nThe computed value is 20.32 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{20.32}.\n#### 20.32", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF031"} {"answer": "0.040", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.257, "old_value": 0.285, "parameter": "WI"}, "distractor_formula_ids": ["CMF020", "CMF018"], "effective_parameters": {"OMS": 298.4, "SI": 0.69, "TI": 0.25, "WI": 0.257}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.003 * OMS", "id": "CMF021", "latex": "OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS", "name": "土壤有机质碳分解量", "name_en": "Soil organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF021_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]", "knowledge_formula_expr": "Formula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "parameters": {"OMS": 298.4, "SI": 0.69, "TI": 0.25, "WI": 0.285}, "problem": "In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil organic matter carbon decomposition.\nUse the verified formula:\n\\[OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\\]\nIn calculation-friendly notation, this is:\nOMSC = WI * SI * TI * 0.003 * OMS.\nSubstitute the relevant values:\nWI=0.257 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2.\nThe computed value is 0.040 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.040}.\n#### 0.040", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF020"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 32.6, "old_value": 29.6, "parameter": "T"}, "distractor_formula_ids": ["CMF028", "CMF044"], "effective_parameters": {"T": 32.6, "Tb": 10.0, "Tmax": 45.0, "Tol": 24.0, "Tou": 34.0}, "family": "growth_yield", "formula": {"expression": "if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Temperature response factor.\nUse the verified formula:\n\\[F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\\]\nIn calculation-friendly notation, this is:\nF_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Potential leaf area index (潜在叶面积指数).\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Potential leaf area index.\nUse the verified formula:\n\\[LAI_P=LAI_0e^{RP\\cdot GDD}\\]\nIn calculation-friendly notation, this is:\nLAIP = LAI0 * exp(RP * GDD).\nSubstitute the relevant values:\nLAI0=0.315 m2/m2, RP=0.0047 d-1, GDD=275.6 C day.\nThe computed value is 1.15 m2/m2.\nTherefore, the final numeric answer is \\boxed{1.15}.\n#### 1.15", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "wrong_formula_id": "CMF043"} {"answer": "10.60", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF057", "CMF059"], "effective_parameters": {"Tbase": 5.1, "Tmean": 15.7}, "family": "environmental_response", "formula": {"expression": "max(Tmean - Tbase, 0)", "id": "CMF056", "latex": "dTT=\\max(T_{mean}-T_{base},0)", "name": "每日有效积温", "name_en": "environmental-response daily thermal time", "precision": 2, "unit": "C day"}, "id": "CMF056_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]", "knowledge_formula_expr": "Formula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_name": "Relevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Maximum linear", "mode": "direct", "parameters": {"Tbase": 5.1, "Tmean": 15.7}, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily thermal time.\nUse the verified formula:\n\\[dTT=\\max(T_{mean}-T_{base},0)\\]\nIn calculation-friendly notation, this is:\ndTT = max(Tmean - Tbase, 0).\nSubstitute the relevant values:\nTmean=15.7 C, Tbase=5.1 C.\nThe computed value is 10.60 C day.\nTherefore, the final numeric answer is \\boxed{10.60}.\n#### 10.60", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF057"} {"answer": "0.44", "benchmark_version": "cropmath-v1", "change": {"new_value": 22.4, "old_value": 20.4, "parameter": "PDT"}, "distractor_formula_ids": ["CMF036", "CMF031"], "effective_parameters": {"PDT": 22.4}, "family": "growth_yield", "formula": {"expression": "if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)", "id": "CMF041", "latex": "PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)", "name": "绿叶分配指数", "name_en": "Green leaf partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF041_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Green leaf partitioning index (绿叶分配指数).\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "parameters": {"PDT": 20.4}, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Green leaf partitioning index (绿叶分配指数).\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Green leaf partitioning index.\nUse the verified formula:\n\\[PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\\]\nIn calculation-friendly notation, this is:\nPIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT).\nSubstitute the relevant values:\nPDT=22.4 physiological days.\nThe computed value is 0.44 fraction.\nTherefore, the final numeric answer is \\boxed{0.44}.\n#### 0.44", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF036"} {"answer": "0.563", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF007", "CMF006"], "effective_parameters": {"Coef_K_Decom": 0.0102, "Coef_K_Trans": 0.152, "F_notill": 1.14, "F_pH": 0.18, "F_temp": 1.209, "F_texture": 0.959, "F_water": 0.532, "Pool": 514.7}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)", "id": "CMF009", "latex": "R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})", "name": "碳库呼吸损失", "name_en": "carbon-nitrogen cycling carbon pool respiration loss", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF009_chain_2_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]", "knowledge_formula_expr": "Formula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "parameters": {"Coef_K_Decom": 0.0102, "Coef_K_Trans": 0.152, "F_notill": 1.14, "F_pH": 0.18, "F_temp": 1.209, "F_texture": 0.959, "F_water": 0.532, "Pool": 514.7}, "problem": "In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling carbon pool respiration loss.\nUse the verified formula:\n\\[R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\\]\nIn calculation-friendly notation, this is:\nRespiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans).\nSubstitute the relevant values:\nPool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction.\nThe computed value is 0.563 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.563}.\n#### 0.563", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "wrong_formula_id": "CMF007"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 24.0, "old_value": 21.8, "parameter": "PDT"}, "distractor_formula_ids": ["CMF049", "CMF029"], "effective_parameters": {"PDT": 24.0, "a": 0.039}, "family": "growth_yield", "formula": {"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))", "id": "CMF030", "latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)", "name": "生理年龄因子", "name_en": "Physiological age factor", "precision": 2, "unit": "fraction"}, "id": "CMF030_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]", "knowledge_formula_expr": "Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "parameters": {"PDT": 21.8, "a": 0.039}, "problem": "Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Physiological age factor.\nUse the verified formula:\n\\[F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\\]\nIn calculation-friendly notation, this is:\nFPA = if PDT<28: 1; else: exp(-a*(PDT-28)).\nSubstitute the relevant values:\nPDT=24.0 physiological days, a=0.039 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF049"} {"answer": "0.16", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF047", "CMF027"], "effective_parameters": {"PDT": 27.6}, "family": "growth_yield", "formula": {"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)", "id": "CMF043", "latex": "PIRO=1-PISH", "name": "地下部分配指数", "name_en": "Root partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF043_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "parameters": {"PDT": 27.6}, "problem": "Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root partitioning index.\nUse the verified formula:\n\\[PIRO=1-PISH\\]\nIn calculation-friendly notation, this is:\nPIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).\nSubstitute the relevant values:\nPDT=27.6 physiological days.\nThe computed value is 0.16 fraction.\nTherefore, the final numeric answer is \\boxed{0.16}.\n#### 0.16", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF047"} {"answer": "0.343", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF040", "CMF042"], "effective_parameters": {"T": 15.0, "TS": 1.19, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "family": "growth_yield", "formula": {"expression": "(((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS", "id": "CMF051", "latex": "RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}", "name": "相对热效应", "name_en": "Relative thermal effectiveness", "precision": 3, "unit": "fraction"}, "id": "CMF051_direct_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]", "knowledge_formula_expr": "Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Beta temperature response", "mode": "direct", "parameters": {"T": 15.0, "TS": 1.19, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "problem": "A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Relative thermal effectiveness.\nUse the verified formula:\n\\[RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\\]\nIn calculation-friendly notation, this is:\nRTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS.\nSubstitute the relevant values:\nT=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless.\nThe computed value is 0.343 fraction.\nTherefore, the final numeric answer is \\boxed{0.343}.\n#### 0.343", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF040"} {"answer": "204.19", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF003", "CMF005"], "effective_parameters": {"Coef_K": 0.6, "Coef_rou": 0.15, "LAI": 5.57, "Radi": 411.2}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)", "id": "CMF006", "latex": "PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}", "name": "冠层吸收光合有效辐射", "name_en": "carbon-nitrogen cycling canopy absorbed PAR", "precision": 2, "unit": "umol/m2/s"}, "id": "CMF006_chain_2_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]", "knowledge_formula_expr": "Formula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "parameters": {"Coef_K": 0.6, "Coef_rou": 0.15, "LAI": 5.57, "Radi": 411.2}, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling canopy absorbed PAR.\nUse the verified formula:\n\\[PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\\]\nIn calculation-friendly notation, this is:\nPAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI).\nSubstitute the relevant values:\nRadi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless.\nThe computed value is 204.19 umol/m2/s.\nTherefore, the final numeric answer is \\boxed{204.19}.\n#### 204.19", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF003"} {"answer": "0.901", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.566, "old_value": 0.629, "parameter": "TI"}, "distractor_formula_ids": ["CMF025", "CMF024"], "effective_parameters": {"OMN": 249.2, "SI": 0.91, "TI": 0.566, "WI": 0.26}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.027 * OMN", "id": "CMF020", "latex": "OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN", "name": "新鲜有机质碳分解量", "name_en": "Fresh organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF020_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]", "knowledge_formula_expr": "Formula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "parameters": {"OMN": 249.2, "SI": 0.91, "TI": 0.629, "WI": 0.26}, "problem": "Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Fresh organic matter carbon decomposition.\nUse the verified formula:\n\\[OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\\]\nIn calculation-friendly notation, this is:\nOMNC = WI * SI * TI * 0.027 * OMN.\nSubstitute the relevant values:\nWI=0.26 fraction, SI=0.91 fraction, TI=0.566 fraction, OMN=249.2 g C/m2.\nThe computed value is 0.901 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.901}.\n#### 0.901", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF025"} {"answer": "1.605", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF055", "CMF056"], "effective_parameters": {"CO2": 739.0, "CO2_RUE": 0.173}, "family": "environmental_response", "formula": {"expression": "if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)", "id": "CMF057", "latex": "F_{CO2}=f(CO2,CO2_{RUE})", "name": "环境响应二氧化碳响应", "name_en": "environmental-response CO2 response", "precision": 3, "unit": "multiplier"}, "id": "CMF057_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]", "knowledge_formula_expr": "Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "parameters": {"CO2": 739.0, "CO2_RUE": 0.173}, "problem": "In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response CO2 response.\nUse the verified formula:\n\\[F_{CO2}=f(CO2,CO2_{RUE})\\]\nIn calculation-friendly notation, this is:\nF_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1).\nSubstitute the relevant values:\nCO2=739.0 ppm, CO2_RUE=0.173 fraction.\nThe computed value is 1.605 fraction.\nTherefore, the final numeric answer is \\boxed{1.605}.\n#### 1.605", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF055"} {"answer": "6.456", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF001"], "effective_parameters": {"Day_Length_hour": 11.99, "F_CO2": 1.14, "F_T": 0.49, "F_Water": 0.43, "Photo_Rate_opt": 51.89}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432", "id": "CMF002", "latex": "GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432", "name": "总初级生产力", "name_en": "carbon-nitrogen cycling gross primary production", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF002_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]", "knowledge_formula_expr": "Formula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "parameters": {"Day_Length_hour": 11.99, "F_CO2": 1.14, "F_T": 0.49, "F_Water": 0.43, "Photo_Rate_opt": 51.89}, "problem": "Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling gross primary production.\nUse the verified formula:\n\\[GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\\]\nIn calculation-friendly notation, this is:\nGPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432.\nSubstitute the relevant values:\nPhoto_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h.\nThe computed value is 6.456 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{6.456}.\n#### 6.456", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "wrong_formula_id": "CMF009"} {"answer": "113.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF061", "CMF062"], "effective_parameters": {"F_CO2": 1.26, "F_Heat": 1.0, "F_Temp": 0.7, "F_Water": 0.65, "RUE": 2.5, "SRAD": 12.13, "fSolar": 0.65}, "family": "environmental_response", "formula": {"expression": "10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)", "id": "CMF055", "latex": "dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)", "name": "环境响应日生物量增量", "name_en": "environmental-response daily biomass increment", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF055_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]", "knowledge_formula_expr": "Formula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "parameters": {"F_CO2": 1.26, "F_Heat": 1.0, "F_Temp": 0.7, "F_Water": 0.65, "RUE": 2.5, "SRAD": 12.13, "fSolar": 0.65}, "problem": "A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response daily biomass increment.\nUse the verified formula:\n\\[dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\\]\nIn calculation-friendly notation, this is:\ndBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat).\nSubstitute the relevant values:\nRUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction.\nThe computed value is 113.00 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{113.00}.\n#### 113.00", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF061"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 17.8, "old_value": 16.2, "parameter": "PDT"}, "distractor_formula_ids": ["CMF042", "CMF041"], "effective_parameters": {"PDT": 17.8, "a": 0.035}, "family": "growth_yield", "formula": {"expression": "if PDT<28: 1; else: exp(-a*(PDT-28))", "id": "CMF030", "latex": "F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)", "name": "生理年龄因子", "name_en": "Physiological age factor", "precision": 2, "unit": "fraction"}, "id": "CMF030_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]", "knowledge_formula_expr": "Formula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "parameters": {"PDT": 16.2, "a": 0.035}, "problem": "In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Physiological age factor.\nUse the verified formula:\n\\[F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\\]\nIn calculation-friendly notation, this is:\nFPA = if PDT<28: 1; else: exp(-a*(PDT-28)).\nSubstitute the relevant values:\nPDT=17.8 physiological days, a=0.035 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF042"} {"answer": "0.94", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF052", "CMF032"], "effective_parameters": {"PDT": 29.3, "Qt": 7.8, "Tmean": 18.4}, "family": "growth_yield", "formula": {"expression": "if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1", "id": "CMF039", "latex": "LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}", "name": "低温结实因子", "name_en": "Low-temperature fertility factor", "precision": 2, "unit": "fraction"}, "id": "CMF039_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Low-temperature fertility factor.\nUse the verified formula:\n\\[LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\\]\nIn calculation-friendly notation, this is:\nLTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1.\nSubstitute the relevant values:\nPDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day.\nThe computed value is 0.94 fraction.\nTherefore, the final numeric answer is \\boxed{0.94}.\n#### 0.94", "split": "dev", "subject": "Temperature stress", "template_id": 0, "wrong_formula_id": "CMF052"} {"answer": "0.922", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF011", "CMF002"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.511, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise water response", "mode": "branch", "parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.511, "Wilt_point": 0.08}, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop water stress factor.\nUse the verified formula:\n\\[F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\\]\nIn calculation-friendly notation, this is:\nF_Water = if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration.\nUse the verified formula:\n\\[RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\\]\nIn calculation-friendly notation, this is:\nRM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10).\nSubstitute the relevant values:\nPDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C.\nThe computed value is 52.19 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{52.19}.\n#### 52.19", "split": "dev", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF039"} {"answer": "0.32", "benchmark_version": "cropmath-v1", "change": {"new_value": 5.6, "old_value": 4.9, "parameter": "PDT"}, "distractor_formula_ids": ["CMF036", "CMF040"], "effective_parameters": {"PDT": 5.6}, "family": "growth_yield", "formula": {"expression": "1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)", "id": "CMF043", "latex": "PIRO=1-PISH", "name": "地下部分配指数", "name_en": "Root partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF043_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "parameters": {"PDT": 4.9}, "problem": "In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Root partitioning index.\nUse the verified formula:\n\\[PIRO=1-PISH\\]\nIn calculation-friendly notation, this is:\nPIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63).\nSubstitute the relevant values:\nPDT=5.6 physiological days.\nThe computed value is 0.32 fraction.\nTherefore, the final numeric answer is \\boxed{0.32}.\n#### 0.32", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF036"} {"answer": "0.000", "benchmark_version": "cropmath-v1", "change": {"new_value": 25.4, "old_value": 23.1, "parameter": "Topt"}, "distractor_formula_ids": ["CMF056", "CMF059"], "effective_parameters": {"Tbase": 4.3, "Tmean": 0.2, "Topt": 25.4}, "family": "environmental_response", "formula": {"expression": "if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)", "id": "CMF060", "latex": "F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)", "name": "环境响应温度响应", "name_en": "environmental-response temperature response", "precision": 3, "unit": "fraction"}, "id": "CMF060_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]", "knowledge_formula_expr": "Formula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "parameters": {"Tbase": 4.3, "Tmean": 0.2, "Topt": 23.1}, "problem": "Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response temperature response.\nUse the verified formula:\n\\[F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\\]\nIn calculation-friendly notation, this is:\nF_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0).\nSubstitute the relevant values:\nTmean=0.2 C, Tbase=4.3 C, Topt=25.4 C.\nThe computed value is 0.000 fraction.\nTherefore, the final numeric answer is \\boxed{0.000}.\n#### 0.000", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF056"} {"answer": "1.100", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF008", "CMF003"], "effective_parameters": {"Temp": 8.8, "is_soil_temp": 0.0}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)", "id": "CMF011", "latex": "F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T", "name": "土壤碳分解温度因子", "name_en": "carbon-nitrogen cycling soil carbon temperature factor", "precision": 3, "unit": "multiplier"}, "id": "CMF011_branch_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]", "knowledge_formula_expr": "Formula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "parameters": {"Temp": 8.8, "is_soil_temp": 0.0}, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling soil carbon temperature factor.\nUse the verified formula:\n\\[F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\\]\nIn calculation-friendly notation, this is:\nF_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1).\nSubstitute the relevant values:\nTemp=8.8 C, is_soil_temp=0.0 0/1.\nThe computed value is 1.100 fraction.\nTherefore, the final numeric answer is \\boxed{1.100}.\n#### 1.100", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "wrong_formula_id": "CMF008"} {"answer": "1.000", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF001", "CMF006"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.288, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Watersystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop water stress factor.\nUse the verified formula:\n\\[F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\\]\nIn calculation-friendly notation, this is:\nF_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)", "id": "CMF057", "latex": "F_{CO2}=f(CO2,CO2_{RUE})", "name": "环境响应二氧化碳响应", "name_en": "environmental-response CO2 response", "precision": 3, "unit": "multiplier"}, "id": "CMF057_direct_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]", "knowledge_formula_expr": "Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "parameters": {"CO2": 630.0, "CO2_RUE": 0.095}, "problem": "Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response CO2 response.\nUse the verified formula:\n\\[F_{CO2}=f(CO2,CO2_{RUE})\\]\nIn calculation-friendly notation, this is:\nF_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1).\nSubstitute the relevant values:\nCO2=630.0 ppm, CO2_RUE=0.095 fraction.\nThe computed value is 1.266 fraction.\nTherefore, the final numeric answer is \\boxed{1.266}.\n#### 1.266", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF056"} {"answer": "0.308", "benchmark_version": "cropmath-v1", "change": {"new_value": 40.0, "old_value": 39.5, "parameter": "T"}, "distractor_formula_ids": ["CMF036", "CMF052"], "effective_parameters": {"T": 40.0, "TS": 1.76, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "family": "growth_yield", "formula": {"expression": "(((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS", "id": "CMF051", "latex": "RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}", "name": "相对热效应", "name_en": "Relative thermal effectiveness", "precision": 3, "unit": "fraction"}, "id": "CMF051_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]", "knowledge_formula_expr": "Formula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "parameters": {"T": 39.5, "TS": 1.76, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Relative thermal effectiveness.\nUse the verified formula:\n\\[RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\\]\nIn calculation-friendly notation, this is:\nRTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS.\nSubstitute the relevant values:\nT=40.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless.\nThe computed value is 0.308 fraction.\nTherefore, the final numeric answer is \\boxed{0.308}.\n#### 0.308", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF036"} {"answer": "0.294", "benchmark_version": "cropmath-v1", "change": {"new_value": 8.2, "old_value": 7.1, "parameter": "PDT"}, "distractor_formula_ids": ["CMF046", "CMF053"], "effective_parameters": {"PDT": 8.2}, "family": "growth_yield", "formula": {"expression": "0.0087 * PDT + 0.2222", "id": "CMF036", "latex": "K=0.0087PDT+0.2222", "name": "冠层消光系数", "name_en": "Canopy extinction coefficient", "precision": 3, "unit": "dimensionless"}, "id": "CMF036_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless", "knowledge_name": "Relevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Linear", "mode": "sensitivity", "parameters": {"PDT": 7.1}, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Canopy extinction coefficient.\nUse the verified formula:\n\\[K=0.0087PDT+0.2222\\]\nIn calculation-friendly notation, this is:\nK = 0.0087 * PDT + 0.2222.\nSubstitute the relevant values:\nPDT=8.2 physiological days.\nThe computed value is 0.294 dimensionless.\nTherefore, the final numeric answer is \\boxed{0.294}.\n#### 0.294", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF046"} {"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 33.8, "old_value": 37.6, "parameter": "T"}, "distractor_formula_ids": ["CMF044", "CMF038"], "effective_parameters": {"T": 33.8, "Tb": 10.0, "Tmax": 45.0, "Tol": 24.0, "Tou": 34.0}, "family": "growth_yield", "formula": {"expression": "if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Temperature response factor.\nUse the verified formula:\n\\[F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\\]\nIn calculation-friendly notation, this is:\nF_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Actual LAI relative growth rate.\nUse the verified formula:\n\\[R=RP\\cdot\\min(NNI,WDF)\\]\nIn calculation-friendly notation, this is:\nR = RP * min(NNI, WDF).\nSubstitute the relevant values:\nRP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction.\nThe computed value is 0.0046 d-1.\nTherefore, the final numeric answer is \\boxed{0.0046}.\n#### 0.0046", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF049"} {"answer": "398.53", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF026", "CMF030"], "effective_parameters": {"DTGA": 502.1, "RG": 61.2, "RM": 62.3, "b": 0.05}, "family": "growth_yield", "formula": {"expression": "(DTGA - RM - RG) / (1 - b)", "id": "CMF033", "latex": "GCR=\\frac{DTGA-RM-RG}{1-b}", "name": "日干物质增量", "name_en": "Daily dry matter increment", "precision": 2, "unit": "kg/ha/day"}, "id": "CMF033_chain_2_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]", "knowledge_formula_expr": "Formula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_name": "Relevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "parameters": {"DTGA": 502.1, "RG": 61.2, "RM": 62.3, "b": 0.05}, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily dry matter increment.\nUse the verified formula:\n\\[GCR=\\frac{DTGA-RM-RG}{1-b}\\]\nIn calculation-friendly notation, this is:\nGCR = (DTGA - RM - RG) / (1 - b).\nSubstitute the relevant values:\nDTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction.\nThe computed value is 398.53 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{398.53}.\n#### 398.53", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF026"} {"answer": "0.792", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.05, "old_value": 0.045, "parameter": "Ratio_N_C_photo_product"}, "distractor_formula_ids": ["CMF011", "CMF006"], "effective_parameters": {"Coef_Plant_TN_uptake": 0.117, "DVI": 0.49, "NPP": 15.84, "N_requirement_max_day": 2.09, "N_shoot": 5.24, "Ratio_N_C_photo_product": 0.05}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)", "id": "CMF004", "latex": "N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)", "name": "作物需氮量", "name_en": "carbon-nitrogen cycling crop nitrogen demand", "precision": 3, "unit": "g N/m2/day"}, "id": "CMF004_sensitivity_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]", "knowledge_formula_expr": "Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise minimum", "mode": "sensitivity", "parameters": {"Coef_Plant_TN_uptake": 0.117, "DVI": 0.49, "NPP": 15.84, "N_requirement_max_day": 2.09, "N_shoot": 5.24, "Ratio_N_C_photo_product": 0.045}, "problem": "Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop nitrogen demand.\nUse the verified formula:\n\\[N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\\]\nIn calculation-friendly notation, this is:\nN_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP).\nSubstitute the relevant values:\nN_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.05 g N/g C, DVI=0.49 dimensionless.\nThe computed value is 0.792 g N/m2/day.\nTherefore, the final numeric answer is \\boxed{0.792}.\n#### 0.792", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "wrong_formula_id": "CMF011"} {"answer": "0.481", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF048", "CMF042"], "effective_parameters": {"BFF": 0.72, "DTE": 0.668, "IE": 0.67, "PDT": 39.5, "RPE": 0.5}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]", "knowledge_formula_expr": "Formula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_name": "Relevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "parameters": {"BFF": 0.72, "DTE": 0.668, "IE": 0.67, "PDT": 39.5, "RPE": 0.5}, "problem": "Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless.\nThe computed value is 0.481 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.481}.\n#### 0.481", "split": "dev", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF048"} {"answer": "11.269", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF020", "CMF023"], "effective_parameters": {"Com": 6.15, "Cr": 88.53, "FEh": 0.601, "TI": 0.715}, "family": "methane_emission", "formula": {"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))", "id": "CMF022", "latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))", "name": "甲烷产生量", "name_en": "Methane production", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF022_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]", "knowledge_formula_expr": "Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "parameters": {"Com": 6.15, "Cr": 88.53, "FEh": 0.601, "TI": 0.715}, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane production.\nUse the verified formula:\n\\[P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\\]\nIn calculation-friendly notation, this is:\nP = max(0, 0.27 * FEh * (TI * Cr + Com)).\nSubstitute the relevant values:\nFEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day.\nThe computed value is 11.269 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{11.269}.\n#### 11.269", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF020"} {"answer": "0.0084", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF033", "CMF051"], "effective_parameters": {"PDT": 6.8}, "family": "growth_yield", "formula": {"expression": "0.0091 - 0.0001 * PDT", "id": "CMF049", "latex": "RM(T_o)=0.0091-0.0001PDT", "name": "最适温度维持呼吸系数", "name_en": "Maintenance respiration coefficient at optimum temperature", "precision": 4, "unit": "d-1"}, "id": "CMF049_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_name": "Relevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Linear decline", "mode": "direct", "parameters": {"PDT": 6.8}, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Maintenance respiration coefficient at optimum temperature.\nUse the verified formula:\n\\[RM(T_o)=0.0091-0.0001PDT\\]\nIn calculation-friendly notation, this is:\nRMTo = 0.0091 - 0.0001 * PDT.\nSubstitute the relevant values:\nPDT=6.8 physiological days.\nThe computed value is 0.0084 d-1.\nTherefore, the final numeric answer is \\boxed{0.0084}.\n#### 0.0084", "split": "dev", "subject": "Respiration", "template_id": 2, "wrong_formula_id": "CMF033"} {"answer": "0.680", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF016", "CMF020"], "effective_parameters": {"OMS": 832.8, "SI": 1.0, "TI": 0.56, "WI": 0.486}, "family": "methane_emission", "formula": {"expression": "WI * SI * TI * 0.003 * OMS", "id": "CMF021", "latex": "OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS", "name": "土壤有机质碳分解量", "name_en": "Soil organic matter carbon decomposition", "precision": 3, "unit": "g C/m2/day"}, "id": "CMF021_direct_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]", "knowledge_formula_expr": "Formula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_name": "Relevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Multiplicative", "mode": "direct", "parameters": {"OMS": 832.8, "SI": 1.0, "TI": 0.56, "WI": 0.486}, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Soil organic matter carbon decomposition.\nUse the verified formula:\n\\[OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\\]\nIn calculation-friendly notation, this is:\nOMSC = WI * SI * TI * 0.003 * OMS.\nSubstitute the relevant values:\nWI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2.\nThe computed value is 0.680 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{0.680}.\n#### 0.680", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF016"} {"answer": "2.690", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.315, "old_value": 1.461, "parameter": "TI"}, "distractor_formula_ids": ["CMF016", "CMF020"], "effective_parameters": {"Com": 3.883, "Cr": 7.84, "FEh": 0.702, "TI": 1.315}, "family": "methane_emission", "formula": {"expression": "max(0, 0.27 * FEh * (TI * Cr + Com))", "id": "CMF022", "latex": "P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))", "name": "甲烷产生量", "name_en": "Methane production", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF022_sensitivity_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]", "knowledge_formula_expr": "Formula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Maximum with chained drivers", "mode": "sensitivity", "parameters": {"Com": 3.883, "Cr": 7.84, "FEh": 0.702, "TI": 1.461}, "problem": "Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Methane production.\nUse the verified formula:\n\\[P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\\]\nIn calculation-friendly notation, this is:\nP = max(0, 0.27 * FEh * (TI * Cr + Com)).\nSubstitute the relevant values:\nFEh=0.702 fraction, TI=1.315 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day.\nThe computed value is 2.690 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{2.690}.\n#### 2.690", "split": "dev", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF016"} {"answer": "0.091", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF010", "CMF005"], "effective_parameters": {"Coef_Plant_TN_uptake": 0.166, "DVI": 0.73, "NPP": 1.55, "N_requirement_max_day": 1.64, "N_shoot": 6.78, "Ratio_N_C_photo_product": 0.059}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)", "id": "CMF004", "latex": "N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)", "name": "作物需氮量", "name_en": "carbon-nitrogen cycling crop nitrogen demand", "precision": 3, "unit": "g N/m2/day"}, "id": "CMF004_direct_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]", "knowledge_formula_expr": "Formula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "parameters": {"Coef_Plant_TN_uptake": 0.166, "DVI": 0.73, "NPP": 1.55, "N_requirement_max_day": 1.64, "N_shoot": 6.78, "Ratio_N_C_photo_product": 0.059}, "problem": "In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling crop nitrogen demand.\nUse the verified formula:\n\\[N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\\]\nIn calculation-friendly notation, this is:\nN_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP).\nSubstitute the relevant values:\nN_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless.\nThe computed value is 0.091 g N/m2/day.\nTherefore, the final numeric answer is \\boxed{0.091}.\n#### 0.091", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 2, "wrong_formula_id": "CMF010"} {"answer": "1.392", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF056", "CMF055"], "effective_parameters": {"CO2": 703.0, "CO2_RUE": 0.112}, "family": "environmental_response", "formula": {"expression": "if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)", "id": "CMF057", "latex": "F_{CO2}=f(CO2,CO2_{RUE})", "name": "环境响应二氧化碳响应", "name_en": "environmental-response CO2 response", "precision": 3, "unit": "multiplier"}, "id": "CMF057_branch_t1_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]", "knowledge_formula_expr": "Formula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 4, "math_type": "Piecewise linear cap", "mode": "branch", "parameters": {"CO2": 703.0, "CO2_RUE": 0.112}, "problem": "For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as environmental-response CO2 response.\nUse the verified formula:\n\\[F_{CO2}=f(CO2,CO2_{RUE})\\]\nIn calculation-friendly notation, this is:\nF_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1).\nSubstitute the relevant values:\nCO2=703.0 ppm, CO2_RUE=0.112 fraction.\nThe computed value is 1.392 fraction.\nTherefore, the final numeric answer is \\boxed{1.392}.\n#### 1.392", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF056"} {"answer": "0.58", "benchmark_version": "cropmath-v1", "change": {"new_value": 40.7, "old_value": 37.0, "parameter": "T"}, "distractor_formula_ids": ["CMF044", "CMF043"], "effective_parameters": {"T": 40.7, "Tb": 10.0, "Tmax": 45.0, "Tol": 24.0, "Tou": 34.0}, "family": "growth_yield", "formula": {"expression": "if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Temperature response factor.\nUse the verified formula:\n\\[F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\\]\nIn calculation-friendly notation, this is:\nF_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Nitrogen use efficiency.\nUse the verified formula:\n\\[NUE=\\frac{GrainYield}{NUptake}\\]\nIn calculation-friendly notation, this is:\nNUE = GrainYield / NUptake.\nSubstitute the relevant values:\nGrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha.\nThe computed value is 90.97 kg grain/kg N.\nTherefore, the final numeric answer is \\boxed{90.97}.\n#### 90.97", "split": "dev", "subject": "Nitrogen", "template_id": 0, "wrong_formula_id": "CMF010"} {"answer": "22.164", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF013", "CMF001"], "effective_parameters": {"Coef_alfa": 19.43, "Coef_beta": 354.5, "Leaf_N": 1.58, "PAR": 920.5}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "(Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)", "id": "CMF007", "latex": "P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}", "name": "最适光合速率", "name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "precision": 3, "unit": "umol CO2/m2/s"}, "id": "CMF007_direct_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]", "knowledge_formula_expr": "Formula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "parameters": {"Coef_alfa": 19.43, "Coef_beta": 354.5, "Leaf_N": 1.58, "PAR": 920.5}, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling optimal photosynthetic rate.\nUse the verified formula:\n\\[P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\\]\nIn calculation-friendly notation, this is:\nPhoto_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta).\nSubstitute the relevant values:\nCoef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2.\nThe computed value is 22.164 umol CO2/m2/s.\nTherefore, the final numeric answer is \\boxed{22.164}.\n#### 22.164", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF013"} {"answer": "1.45", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.81, "old_value": 0.9, "parameter": "beta"}, "distractor_formula_ids": ["CMF054", "CMF041"], "effective_parameters": {"Cx": 593.0, "beta": 0.81}, "family": "growth_yield", "formula": {"expression": "1 + beta * ln(Cx / 340)", "id": "CMF029", "latex": "F_{CO2}=1+\\beta\\ln(C_x/340)", "name": "二氧化碳影响因子", "name_en": "CO2 response factor", "precision": 2, "unit": "multiplier"}, "id": "CMF029_sensitivity_t0_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]", "knowledge_formula_expr": "Formula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "parameters": {"Cx": 593.0, "beta": 0.9}, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as CO2 response factor.\nUse the verified formula:\n\\[F_{CO2}=1+\\beta\\ln(C_x/340)\\]\nIn calculation-friendly notation, this is:\nFCO2 = 1 + beta * ln(Cx / 340).\nSubstitute the relevant values:\nCx=593.0 ppm, beta=0.81 dimensionless.\nThe computed value is 1.45 fraction.\nTherefore, the final numeric answer is \\boxed{1.45}.\n#### 1.45", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF054"} {"answer": "8.457", "benchmark_version": "cropmath-v1", "change": {"new_value": 15.106, "old_value": 16.784, "parameter": "P"}, "distractor_formula_ids": ["CMF023", "CMF019"], "effective_parameters": {"CH4RiceEfC": 0.468, "Ebl": 2.514, "P": 15.106, "W": 6006.3, "Wmax": 11997.5}, "family": "methane_emission", "formula": {"expression": "Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25", "id": "CMF017", "latex": "E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}", "name": "甲烷总排放", "name_en": "Total methane emission", "precision": 3, "unit": "g CH4/m2/day"}, "id": "CMF017_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]", "knowledge_formula_expr": "Formula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_name": "Relevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Chained sum", "mode": "sensitivity", "parameters": {"CH4RiceEfC": 0.468, "Ebl": 2.514, "P": 16.784, "W": 6006.3, "Wmax": 11997.5}, "problem": "In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Total methane emission.\nUse the verified formula:\n\\[E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\\]\nIn calculation-friendly notation, this is:\nE = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25.\nSubstitute the relevant values:\nEbl=2.514 g CH4/m2/day, P=15.106 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha.\nThe computed value is 8.457 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{8.457}.\n#### 8.457", "split": "dev", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF023"} {"answer": "0.241", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF039", "CMF032"], "effective_parameters": {"BFF": 0.88, "DTE": 0.274, "IE": 0.91, "PDT": 53.3, "RPE": 0.94}, "family": "growth_yield", "formula": {"expression": "if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0", "id": "CMF027", "latex": "DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF", "name": "每日生理效应", "name_en": "Daily physiological effectiveness", "precision": 3, "unit": "physiological days/day"}, "id": "CMF027_branch_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=Tsystem\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Daily physiological effectiveness.\nUse the verified formula:\n\\[DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\\]\nIn calculation-friendly notation, this is:\nDPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0.\nSubstitute the relevant values:\nPDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless.\nThe computed value is 0.241 physiological days/day.\nTherefore, the final numeric answer is \\boxed{0.241}.\n#### 0.241", "split": "dev", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF039"} {"answer": "39.02", "benchmark_version": "cropmath-v1", "change": {"new_value": 100.4, "old_value": 111.5, "parameter": "NUptake"}, "distractor_formula_ids": ["CMF005", "CMF001"], "effective_parameters": {"GrainYield": 3918.0, "NUptake": 100.4}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "GrainYield / NUptake", "id": "CMF003", "latex": "NUE=\\frac{GrainYield}{NUptake}", "name": "氮利用效率", "name_en": "Nitrogen use efficiency", "precision": 2, "unit": "kg grain/kg N"}, "id": "CMF003_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]", "knowledge_formula_expr": "Formula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_name": "Relevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Ratio", "mode": "sensitivity", "parameters": {"GrainYield": 3918.0, "NUptake": 111.5}, "problem": "A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Nitrogen use efficiency.\nUse the verified formula:\n\\[NUE=\\frac{GrainYield}{NUptake}\\]\nIn calculation-friendly notation, this is:\nNUE = GrainYield / NUptake.\nSubstitute the relevant values:\nGrainYield=3918.0 kg/ha, NUptake=100.4 kg N/ha.\nThe computed value is 39.02 kg grain/kg N.\nTherefore, the final numeric answer is \\boxed{39.02}.\n#### 39.02", "split": "dev", "subject": "Nitrogen", "template_id": 0, "wrong_formula_id": "CMF005"} {"answer": "0.71", "benchmark_version": "cropmath-v1", "change": {"new_value": 8.3, "old_value": 9.2, "parameter": "PDT"}, "distractor_formula_ids": ["CMF045", "CMF037"], "effective_parameters": {"PDT": 8.3}, "family": "growth_yield", "formula": {"expression": "-0.0000842 * PDT^2 + 0.01 * PDT + 0.63", "id": "CMF044", "latex": "PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63", "name": "地上部分配指数", "name_en": "Shoot partitioning index", "precision": 2, "unit": "fraction"}, "id": "CMF044_sensitivity_t2_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]", "knowledge_formula_expr": "Formula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Shoot partitioning index (地上部分配指数).\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "parameters": {"PDT": 9.2}, "problem": "In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Shoot partitioning index (地上部分配指数).\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Shoot partitioning index.\nUse the verified formula:\n\\[PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\\]\nIn calculation-friendly notation, this is:\nPISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63.\nSubstitute the relevant values:\nPDT=8.3 physiological days.\nThe computed value is 0.71 fraction.\nTherefore, the final numeric answer is \\boxed{0.71}.\n#### 0.71", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF045"} {"answer": "0.853", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.522, "old_value": 0.58, "parameter": "WaterC"}, "distractor_formula_ids": ["CMF021", "CMF019"], "effective_parameters": {"WaterC": 0.522}, "family": "methane_emission", "formula": {"expression": "0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)", "id": "CMF025", "latex": "WI=0.49e^{3.88W_c-5.4W_c^2}", "name": "水分分解因子", "name_en": "Water-content decomposition factor", "precision": 3, "unit": "fraction"}, "id": "CMF025_sensitivity_t0_0000", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]", "knowledge_formula_expr": "Formula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "parameters": {"WaterC": 0.58}, "problem": "A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Water-content decomposition factor.\nUse the verified formula:\n\\[WI=0.49e^{3.88W_c-5.4W_c^2}\\]\nIn calculation-friendly notation, this is:\nWI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2).\nSubstitute the relevant values:\nWaterC=0.522 fraction.\nThe computed value is 0.853 fraction.\nTherefore, the final numeric answer is \\boxed{0.853}.\n#### 0.853", "split": "dev", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF021"} {"answer": "21.82", "benchmark_version": "cropmath-v1", "change": {"new_value": 3500.0, "old_value": 3725.0, "parameter": "GrainYield"}, "distractor_formula_ids": ["CMF011", "CMF012"], "effective_parameters": {"GrainYield": 3500.0, "NUptake": 160.4}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "GrainYield / NUptake", "id": "CMF003", "latex": "NUE=\\frac{GrainYield}{NUptake}", "name": "氮利用效率", "name_en": "Nitrogen use efficiency", "precision": 2, "unit": "kg grain/kg N"}, "id": "CMF003_sensitivity_t2_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]", "knowledge_formula_expr": "Formula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_name": "Relevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Ratio", "mode": "sensitivity", "parameters": {"GrainYield": 3725.0, "NUptake": 160.4}, "problem": "In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Nitrogen use efficiency.\nUse the verified formula:\n\\[NUE=\\frac{GrainYield}{NUptake}\\]\nIn calculation-friendly notation, this is:\nNUE = GrainYield / NUptake.\nSubstitute the relevant values:\nGrainYield=3500.0 kg/ha, NUptake=160.4 kg N/ha.\nThe computed value is 21.82 kg grain/kg N.\nTherefore, the final numeric answer is \\boxed{21.82}.\n#### 21.82", "split": "dev", "subject": "Nitrogen", "template_id": 2, "wrong_formula_id": "CMF011"} {"answer": "1.520", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF010", "CMF013"], "effective_parameters": {"CO2": 662.0, "Coef_B_CO2": 0.78}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "1 + Coef_B_CO2 * ln(CO2 / 340)", "id": "CMF001", "latex": "F_{CO2}=1+B\\ln(CO2/340)", "name": "作物二氧化碳施肥因子", "name_en": "carbon-nitrogen cycling CO2 fertilization factor", "precision": 3, "unit": "multiplier"}, "id": "CMF001_direct_t1_0001", "is_gold": false, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_domain": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "Formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]", "knowledge_formula_expr": "Formula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_name": "Relevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction", "knowledge_wrong": "A retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 1, "math_type": "Logarithmic", "mode": "direct", "parameters": {"CO2": 662.0, "Coef_B_CO2": 0.78}, "problem": "Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompts": {"C": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_distractor": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_domain": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_formula": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_name": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "K_wrong": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as carbon-nitrogen cycling CO2 fertilization factor.\nUse the verified formula:\n\\[F_{CO2}=1+B\\ln(CO2/340)\\]\nIn calculation-friendly notation, this is:\nF_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340).\nSubstitute the relevant values:\nCO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless.\nThe computed value is 1.520 fraction.\nTherefore, the final numeric answer is \\boxed{1.520}.\n#### 1.520", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "wrong_formula_id": "CMF010"}