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+{"answer": "0.613", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF056", "CMF062"], "effective_parameters": {"Tbase": 8.0, "Tmean": 19.4, "Topt": 26.6}, "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_0001", "is_gold": true, "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 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 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": 2, "math_type": "Piecewise linear", "mode": "direct", "parameters": {"Tbase": 8.0, "Tmean": 19.4, "Topt": 26.6}, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=19.4 C, Tbase=8.0 C, Topt=26.6 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=19.4 C, Tbase=8.0 C, Topt=26.6 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 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 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 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 temperature response is computed as part of environmental-response formula calculation. With Tmean=19.4 C, Tbase=8.0 C, Topt=26.6 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=19.4 C, Tbase=8.0 C, Topt=26.6 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=19.4 C, Tbase=8.0 C, Topt=26.6 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=19.4 C, Tbase=8.0 C, Topt=26.6 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 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 temperature response is computed as part of environmental-response formula calculation. With Tmean=19.4 C, Tbase=8.0 C, Topt=26.6 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=19.4 C, Tbase=8.0 C, Topt=26.6 C.\nThe computed value is 0.613 fraction.\nTherefore, the final numeric answer is \\boxed{0.613}.\n#### 0.613", "split": "gold", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF056"}
+{"answer": "0.021", "benchmark_version": "cropmath-v1", "change": {"new_value": 13.4, "old_value": 14.9, "parameter": "t_soil"}, "distractor_formula_ids": ["CMF019", "CMF015"], "effective_parameters": {"P": 7.27, "Wr": 629.9, "t_soil": 13.4}, "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_t0_0001", "is_gold": true, "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: 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 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: 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: 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": "Logarithmic minimum", "mode": "sensitivity", "parameters": {"P": 7.27, "Wr": 629.9, "t_soil": 14.9}, "problem": "A methane calculation involves sensitivity analysis. The current measurements are P=7.27 g CH4/m2/day, t_soil=14.9 C, Wr=629.9 kg/ha. A field technician asks what the methane ebullition emission would become if t_soil changed to 13.4. Report the new 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: A methane calculation involves sensitivity analysis. The current measurements are P=7.27 g CH4/m2/day, t_soil=14.9 C, Wr=629.9 kg/ha. A field technician asks what the methane ebullition emission would become if t_soil changed to 13.4. Report the new 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: 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 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: A methane calculation involves sensitivity analysis. The current measurements are P=7.27 g CH4/m2/day, t_soil=14.9 C, Wr=629.9 kg/ha. A field technician asks what the methane ebullition emission would become if t_soil changed to 13.4. Report the new 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: A methane calculation involves sensitivity analysis. The current measurements are P=7.27 g CH4/m2/day, t_soil=14.9 C, Wr=629.9 kg/ha. A field technician asks what the methane ebullition emission would become if t_soil changed to 13.4. Report the new 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: A methane calculation involves sensitivity analysis. The current measurements are P=7.27 g CH4/m2/day, t_soil=14.9 C, Wr=629.9 kg/ha. A field technician asks what the methane ebullition emission would become if t_soil changed to 13.4. Report the new 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: A methane calculation involves sensitivity analysis. The current measurements are P=7.27 g CH4/m2/day, t_soil=14.9 C, Wr=629.9 kg/ha. A field technician asks what the methane ebullition emission would become if t_soil changed to 13.4. Report the new 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: 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 P=7.27 g CH4/m2/day, t_soil=14.9 C, Wr=629.9 kg/ha. A field technician asks what the methane ebullition emission would become if t_soil changed to 13.4. Report the new 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=7.27 g CH4/m2/day, t_soil=13.4 C, Wr=629.9 kg/ha.\nThe computed value is 0.021 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.021}.\n#### 0.021", "split": "gold", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF019"}
+{"answer": "0.938", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF012", "CMF003"], "effective_parameters": {"pH": 6.09}, "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_direct_t2_0001", "is_gold": true, "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 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: 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 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_formula": "Formula 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]", "knowledge_formula_expr": "Formula: 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", "knowledge_name": "Relevant 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", "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": 1, "math_type": "Logistic", "mode": "direct", "parameters": {"pH": 6.09}, "problem": "In a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling soil pH decomposition factor is computed as part of carbon-nitrogen cycling soil carbon modeling. With pH=6.09 pH, 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_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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling soil pH decomposition factor is computed as part of carbon-nitrogen cycling soil carbon modeling. With pH=6.09 pH, 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 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 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: 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 soil carbon, the carbon-nitrogen cycling soil pH decomposition factor is computed as part of carbon-nitrogen cycling soil carbon modeling. With pH=6.09 pH, 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 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 soil carbon, the carbon-nitrogen cycling soil pH decomposition factor is computed as part of carbon-nitrogen cycling soil carbon modeling. With pH=6.09 pH, 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 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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling soil pH decomposition factor is computed as part of carbon-nitrogen cycling soil carbon modeling. With pH=6.09 pH, 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 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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling soil pH decomposition factor is computed as part of carbon-nitrogen cycling soil carbon modeling. With pH=6.09 pH, 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 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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling soil pH decomposition factor is computed as part of carbon-nitrogen cycling soil carbon modeling. With pH=6.09 pH, determine the 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=6.09 pH.\nThe computed value is 0.938 fraction.\nTherefore, the final numeric answer is \\boxed{0.938}.\n#### 0.938", "split": "gold", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "wrong_formula_id": "CMF012"}
+{"answer": "40.55", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF029", "CMF051"], "effective_parameters": {"ABIOMASS": 14935.0, "PDT": 55.4, "Q10": 1.78, "T": 25.3, "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_chain_2_t0_0001", "is_gold": true, "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 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 (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: 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": "Product with Q10 temperature response", "mode": "chain_2", "parameters": {"ABIOMASS": 14935.0, "PDT": 55.4, "Q10": 1.78, "T": 25.3, "To": 30.0}, "problem": "A respiration module estimates Maintenance respiration using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=55.4 physiological days, ABIOMASS=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C. Compute the maintenance respiration and 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: 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: A respiration module estimates Maintenance respiration using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=55.4 physiological days, ABIOMASS=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C. Compute the maintenance respiration and 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: 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 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 (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 module estimates Maintenance respiration using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=55.4 physiological days, ABIOMASS=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C. Compute the maintenance respiration and 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: 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 module estimates Maintenance respiration using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=55.4 physiological days, ABIOMASS=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C. Compute the maintenance respiration and 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: 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: A respiration module estimates Maintenance respiration using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=55.4 physiological days, ABIOMASS=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C. Compute the maintenance respiration and 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: 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: A respiration module estimates Maintenance respiration using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=55.4 physiological days, ABIOMASS=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C. Compute the maintenance respiration and 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: 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 respiration module estimates Maintenance respiration using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=55.4 physiological days, ABIOMASS=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C. Compute the maintenance respiration and give 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=14935.0 kg/ha, Q10=1.78 dimensionless, T=25.3 C, To=30.0 C.\nThe computed value is 40.55 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{40.55}.\n#### 40.55", "split": "gold", "subject": "Respiration", "template_id": 0, "wrong_formula_id": "CMF029"}
+{"answer": "0.506", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF059", "CMF060"], "effective_parameters": {"ExtremeT": 43.2, "MaxT": 35.3, "Tmax": 39.2}, "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_branch_t0_0001", "is_gold": true, "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 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": 4, "math_type": "Piecewise linear stress", "mode": "branch", "parameters": {"ExtremeT": 43.2, "MaxT": 35.3, "Tmax": 39.2}, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmax=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C. Determine the simple heat stress 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_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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmax=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C. Determine the simple heat stress 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 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: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmax=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C. Determine the simple heat stress 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 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 Tmax=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C. Determine the simple heat stress 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 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmax=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C. Determine the simple heat stress 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 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmax=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C. Determine the simple heat stress 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 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmax=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C. Determine the simple heat stress response and report 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=39.2 C, MaxT=35.3 C, ExtremeT=43.2 C.\nThe computed value is 0.506 fraction.\nTherefore, the final numeric answer is \\boxed{0.506}.\n#### 0.506", "split": "gold", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF059"}
+{"answer": "2.380", "benchmark_version": "cropmath-v1", "change": {"new_value": 20.4, "old_value": 17.7, "parameter": "Temp"}, "distractor_formula_ids": ["CMF004", "CMF014"], "effective_parameters": {"Temp": 20.4, "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_sensitivity_t0_0000", "is_gold": true, "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 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\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 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.", "level": 3, "math_type": "Q10 temperature response", "mode": "sensitivity", "parameters": {"Temp": 17.7, "is_soil_temp": 0.0}, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Temp=17.7 C, is_soil_temp=0.0 0/1. A field technician asks what the carbon-nitrogen cycling soil carbon temperature factor would become if Temp changed to 20.4. 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_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 calculation involves sensitivity analysis. The current measurements are Temp=17.7 C, is_soil_temp=0.0 0/1. A field technician asks what the carbon-nitrogen cycling soil carbon temperature factor would become if Temp changed to 20.4. 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: 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\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 calculation involves sensitivity analysis. The current measurements are Temp=17.7 C, is_soil_temp=0.0 0/1. A field technician asks what the carbon-nitrogen cycling soil carbon temperature factor would become if Temp changed to 20.4. 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 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 calculation involves sensitivity analysis. The current measurements are Temp=17.7 C, is_soil_temp=0.0 0/1. A field technician asks what the carbon-nitrogen cycling soil carbon temperature factor would become if Temp changed to 20.4. 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 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 calculation involves sensitivity analysis. The current measurements are Temp=17.7 C, is_soil_temp=0.0 0/1. A field technician asks what the carbon-nitrogen cycling soil carbon temperature factor would become if Temp changed to 20.4. 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 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 calculation involves sensitivity analysis. The current measurements are Temp=17.7 C, is_soil_temp=0.0 0/1. A field technician asks what the carbon-nitrogen cycling soil carbon temperature factor would become if Temp changed to 20.4. 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 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 soil carbon calculation involves sensitivity analysis. The current measurements are Temp=17.7 C, is_soil_temp=0.0 0/1. A field technician asks what the carbon-nitrogen cycling soil carbon temperature factor would become if Temp changed to 20.4. 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 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=20.4 C, is_soil_temp=0.0 0/1.\nThe computed value is 2.380 fraction.\nTherefore, the final numeric answer is \\boxed{2.380}.\n#### 2.380", "split": "gold", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "wrong_formula_id": "CMF004"}
+{"answer": "69.61", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF005", "CMF013"], "effective_parameters": {"GrainYield": 5722.0, "NUptake": 82.2}, "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_0000", "is_gold": true, "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: 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 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: 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": 1, "math_type": "Ratio", "mode": "direct", "parameters": {"GrainYield": 5722.0, "NUptake": 82.2}, "problem": "Given GrainYield=5722.0 kg/ha, NUptake=82.2 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=5722.0 kg/ha, NUptake=82.2 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 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: 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 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 GrainYield=5722.0 kg/ha, NUptake=82.2 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=5722.0 kg/ha, NUptake=82.2 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=5722.0 kg/ha, NUptake=82.2 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=5722.0 kg/ha, NUptake=82.2 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 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 GrainYield=5722.0 kg/ha, NUptake=82.2 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=5722.0 kg/ha, NUptake=82.2 kg N/ha.\nThe computed value is 69.61 kg grain/kg N.\nTherefore, the final numeric answer is \\boxed{69.61}.\n#### 69.61", "split": "gold", "subject": "Nitrogen", "template_id": 1, "wrong_formula_id": "CMF005"}
+{"answer": "0.974", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.14, "old_value": 1.27, "parameter": "BD"}, "distractor_formula_ids": ["CMF005", "CMF004"], "effective_parameters": {"BD": 1.14, "Saturate_Water": 0.48, "Soil_Water": 0.459}, "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_t1_0001", "is_gold": true, "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 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 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 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": 0.459}, "problem": "Given current measurements Soil_Water=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3 in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling soil carbon water factor be if BD were adjusted to 1.14? 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_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: Given current measurements Soil_Water=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3 in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling soil carbon water factor be if BD were adjusted to 1.14? 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: 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 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 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 Soil_Water=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3 in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling soil carbon water factor be if BD were adjusted to 1.14? 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: 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 Soil_Water=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3 in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling soil carbon water factor be if BD were adjusted to 1.14? 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: 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: Given current measurements Soil_Water=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3 in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling soil carbon water factor be if BD were adjusted to 1.14? 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: 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: Given current measurements Soil_Water=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3 in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling soil carbon water factor be if BD were adjusted to 1.14? 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: 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 Soil_Water=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3 in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling soil carbon water factor be if BD were adjusted to 1.14? Provide the 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=0.459 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.14 g/cm3.\nThe computed value is 0.974 fraction.\nTherefore, the final numeric answer is \\boxed{0.974}.\n#### 0.974", "split": "gold", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "wrong_formula_id": "CMF005"}
+{"answer": "0.104", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF021", "CMF023"], "effective_parameters": {"Eh": 49.9}, "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_t0_0001", "is_gold": true, "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: 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: 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: 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": "Piecewise exponential", "mode": "direct", "parameters": {"Eh": 49.9}, "problem": "A methane calculation estimates Redox potential factor. The measured inputs are Eh=49.9 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=49.9 mV. What is the redox potential 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: 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: 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 Redox potential factor. The measured inputs are Eh=49.9 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=49.9 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=49.9 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=49.9 mV. What is the redox potential 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 Redox potential factor. The measured inputs are Eh=49.9 mV. What is the redox potential factor? Give 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=49.9 mV.\nThe computed value is 0.104 fraction.\nTherefore, the final numeric answer is \\boxed{0.104}.\n#### 0.104", "split": "gold", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF021"}
+{"answer": "4925.7", "benchmark_version": "cropmath-v1", "change": {"new_value": 5412.9, "old_value": 4706.9, "parameter": "AWSP"}, "distractor_formula_ids": ["CMF032", "CMF028"], "effective_parameters": {"AWSP": 5412.9, "grain_fraction": 0.91}, "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_sensitivity_t2_0000", "is_gold": true, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\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 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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: 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_formula": "Formula 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]", "knowledge_formula_expr": "Formula: 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", "knowledge_name": "Relevant 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", "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<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Product", "mode": "sensitivity", "parameters": {"AWSP": 4706.9, "grain_fraction": 0.91}, "problem": "In yield formation, we analyze how Dry grain yield responds to parameter changes. Baseline: AWSP=4706.9 kg/ha, grain_fraction=0.91 fraction. After changing AWSP from 4706.9 to 5412.9, compute the new Dry grain yield. Give 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: 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: In yield formation, we analyze how Dry grain yield responds to parameter changes. Baseline: AWSP=4706.9 kg/ha, grain_fraction=0.91 fraction. After changing AWSP from 4706.9 to 5412.9, compute the new Dry grain yield. Give 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: 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 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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: In yield formation, we analyze how Dry grain yield responds to parameter changes. Baseline: AWSP=4706.9 kg/ha, grain_fraction=0.91 fraction. After changing AWSP from 4706.9 to 5412.9, compute the new Dry grain yield. Give 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: 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: In yield formation, we analyze how Dry grain yield responds to parameter changes. Baseline: AWSP=4706.9 kg/ha, grain_fraction=0.91 fraction. After changing AWSP from 4706.9 to 5412.9, compute the new Dry grain yield. Give 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: 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: In yield formation, we analyze how Dry grain yield responds to parameter changes. Baseline: AWSP=4706.9 kg/ha, grain_fraction=0.91 fraction. After changing AWSP from 4706.9 to 5412.9, compute the new Dry grain yield. Give 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: 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: In yield formation, we analyze how Dry grain yield responds to parameter changes. Baseline: AWSP=4706.9 kg/ha, grain_fraction=0.91 fraction. After changing AWSP from 4706.9 to 5412.9, compute the new Dry grain yield. Give 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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In yield formation, we analyze how Dry grain yield responds to parameter changes. Baseline: AWSP=4706.9 kg/ha, grain_fraction=0.91 fraction. After changing AWSP from 4706.9 to 5412.9, compute the new Dry grain yield. Give the numeric value 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=5412.9 kg/ha, grain_fraction=0.91 fraction.\nThe computed value is 4925.7 kg/ha.\nTherefore, the final numeric answer is \\boxed{4925.7}.\n#### 4925.7", "split": "gold", "subject": "Yield formation", "template_id": 2, "wrong_formula_id": "CMF032"}
+{"answer": "0.0027", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.0045, "old_value": 0.0039, "parameter": "RP"}, "distractor_formula_ids": ["CMF041", "CMF032"], "effective_parameters": {"NNI": 0.68, "RP": 0.0045, "WDF": 0.59}, "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_t0_0000", "is_gold": true, "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<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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: 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: 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": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "parameters": {"NNI": 0.68, "RP": 0.0039, "WDF": 0.59}, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are RP=0.0039 d-1, NNI=0.68 fraction, WDF=0.59 fraction. A field technician asks what the actual lai relative growth rate would become if RP changed to 0.0045. 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: 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: A leaf area growth calculation involves sensitivity analysis. The current measurements are RP=0.0039 d-1, NNI=0.68 fraction, WDF=0.59 fraction. A field technician asks what the actual lai relative growth rate would become if RP changed to 0.0045. 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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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: 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 leaf area growth calculation involves sensitivity analysis. The current measurements are RP=0.0039 d-1, NNI=0.68 fraction, WDF=0.59 fraction. A field technician asks what the actual lai relative growth rate would become if RP changed to 0.0045. 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: 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 leaf area growth calculation involves sensitivity analysis. The current measurements are RP=0.0039 d-1, NNI=0.68 fraction, WDF=0.59 fraction. A field technician asks what the actual lai relative growth rate would become if RP changed to 0.0045. 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: 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: A leaf area growth calculation involves sensitivity analysis. The current measurements are RP=0.0039 d-1, NNI=0.68 fraction, WDF=0.59 fraction. A field technician asks what the actual lai relative growth rate would become if RP changed to 0.0045. 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: 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: A leaf area growth calculation involves sensitivity analysis. The current measurements are RP=0.0039 d-1, NNI=0.68 fraction, WDF=0.59 fraction. A field technician asks what the actual lai relative growth rate would become if RP changed to 0.0045. 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: 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 leaf area growth calculation involves sensitivity analysis. The current measurements are RP=0.0039 d-1, NNI=0.68 fraction, WDF=0.59 fraction. A field technician asks what the actual lai relative growth rate would become if RP changed to 0.0045. Report the new 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=0.0045 d-1, NNI=0.68 fraction, WDF=0.59 fraction.\nThe computed value is 0.0027 d-1.\nTherefore, the final numeric answer is \\boxed{0.0027}.\n#### 0.0027", "split": "gold", "subject": "Leaf area growth", "template_id": 0, "wrong_formula_id": "CMF041"}
+{"answer": "0.314", "benchmark_version": "cropmath-v1", "change": {"new_value": 17.3, "old_value": 19.2, "parameter": "t_soil"}, "distractor_formula_ids": ["CMF020", "CMF015"], "effective_parameters": {"Q10": 2.49, "t_soil": 17.3}, "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_t1_0001", "is_gold": true, "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: 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 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": 2.49, "t_soil": 19.2}, "problem": "Given current measurements Q10=2.49 dimensionless, t_soil=19.2 C in methane, what would the Methane temperature index be if t_soil were adjusted to 17.3? 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: 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 current measurements Q10=2.49 dimensionless, t_soil=19.2 C in methane, what would the Methane temperature index be if t_soil were adjusted to 17.3? 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: 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 Q10=2.49 dimensionless, t_soil=19.2 C in methane, what would the Methane temperature index be if t_soil were adjusted to 17.3? 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: 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 Q10=2.49 dimensionless, t_soil=19.2 C in methane, what would the Methane temperature index be if t_soil were adjusted to 17.3? 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: 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 current measurements Q10=2.49 dimensionless, t_soil=19.2 C in methane, what would the Methane temperature index be if t_soil were adjusted to 17.3? 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: 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 current measurements Q10=2.49 dimensionless, t_soil=19.2 C in methane, what would the Methane temperature index be if t_soil were adjusted to 17.3? 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: 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 Q10=2.49 dimensionless, t_soil=19.2 C in methane, what would the Methane temperature index be if t_soil were adjusted to 17.3? Provide 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=2.49 dimensionless, t_soil=17.3 C.\nThe computed value is 0.314 fraction.\nTherefore, the final numeric answer is \\boxed{0.314}.\n#### 0.314", "split": "gold", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF020"}
+{"answer": "0.74", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF044", "CMF053"], "effective_parameters": {"P": 11.46, "PS": 0.279, "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_0001", "is_gold": true, "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: 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: 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: 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: 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": 4, "math_type": "Piecewise quadratic", "mode": "branch", "parameters": {"P": 11.46, "PS": 0.279, "Po": 10.5}, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=11.46 h, Po=10.5 h, PS=0.279 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=11.46 h, Po=10.5 h, PS=0.279 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: 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: 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: 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: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=11.46 h, Po=10.5 h, PS=0.279 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=11.46 h, Po=10.5 h, PS=0.279 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=11.46 h, Po=10.5 h, PS=0.279 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=11.46 h, Po=10.5 h, PS=0.279 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: 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: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=11.46 h, Po=10.5 h, PS=0.279 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=11.46 h, Po=10.5 h, PS=0.279 dimensionless.\nThe computed value is 0.74 fraction.\nTherefore, the final numeric answer is \\boxed{0.74}.\n#### 0.74", "split": "gold", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF044"}
+{"answer": "0.313", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF016", "CMF020"], "effective_parameters": {"Q10": 2.96, "t_soil": 19.3}, "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_t0_0000", "is_gold": true, "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: 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: 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": "Q10 with temperature clipping", "mode": "direct", "parameters": {"Q10": 2.96, "t_soil": 19.3}, "problem": "A methane calculation estimates Methane temperature index. The measured inputs are Q10=2.96 dimensionless, t_soil=19.3 C. What is the methane temperature 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: 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 estimates Methane temperature index. The measured inputs are Q10=2.96 dimensionless, t_soil=19.3 C. What is the methane temperature 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: 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: 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 estimates Methane temperature index. The measured inputs are Q10=2.96 dimensionless, t_soil=19.3 C. What is the methane temperature 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: 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 Methane temperature index. The measured inputs are Q10=2.96 dimensionless, t_soil=19.3 C. What is the methane temperature 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: 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 estimates Methane temperature index. The measured inputs are Q10=2.96 dimensionless, t_soil=19.3 C. What is the methane temperature 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: 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 estimates Methane temperature index. The measured inputs are Q10=2.96 dimensionless, t_soil=19.3 C. What is the methane temperature 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: 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 Methane temperature index. The measured inputs are Q10=2.96 dimensionless, t_soil=19.3 C. What is the methane temperature index? Give 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=2.96 dimensionless, t_soil=19.3 C.\nThe computed value is 0.313 fraction.\nTherefore, the final numeric answer is \\boxed{0.313}.\n#### 0.313", "split": "gold", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF016"}
+{"answer": "34.785", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.25, "old_value": 1.09, "parameter": "F_CO2"}, "distractor_formula_ids": ["CMF006", "CMF011"], "effective_parameters": {"Day_Length_hour": 14.84, "F_CO2": 1.25, "F_T": 0.78, "F_Water": 0.96, "Photo_Rate_opt": 57.97}, "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_t1_0000", "is_gold": true, "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 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 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 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": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "parameters": {"Day_Length_hour": 14.84, "F_CO2": 1.09, "F_T": 0.78, "F_Water": 0.96, "Photo_Rate_opt": 57.97}, "problem": "Given current measurements Photo_Rate_opt=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.09 fraction, Day_Length_hour=14.84 h in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling gross primary production be if F_CO2 were adjusted to 1.25? 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: 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 current measurements Photo_Rate_opt=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.09 fraction, Day_Length_hour=14.84 h in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling gross primary production be if F_CO2 were adjusted to 1.25? 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 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 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 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 current measurements Photo_Rate_opt=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.09 fraction, Day_Length_hour=14.84 h in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling gross primary production be if F_CO2 were adjusted to 1.25? 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 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 current measurements Photo_Rate_opt=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.09 fraction, Day_Length_hour=14.84 h in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling gross primary production be if F_CO2 were adjusted to 1.25? 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 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 current measurements Photo_Rate_opt=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.09 fraction, Day_Length_hour=14.84 h in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling gross primary production be if F_CO2 were adjusted to 1.25? 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 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 current measurements Photo_Rate_opt=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.09 fraction, Day_Length_hour=14.84 h in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling gross primary production be if F_CO2 were adjusted to 1.25? 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 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 Photo_Rate_opt=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.09 fraction, Day_Length_hour=14.84 h in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling gross primary production be if F_CO2 were adjusted to 1.25? 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 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=57.97 umol CO2/m2/s, F_T=0.78 fraction, F_Water=0.96 fraction, F_CO2=1.25 fraction, Day_Length_hour=14.84 h.\nThe computed value is 34.785 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{34.785}.\n#### 34.785", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "wrong_formula_id": "CMF006"}
+{"answer": "0.91", "benchmark_version": "cropmath-v1", "change": {"new_value": 45.1, "old_value": 50.1, "parameter": "PDT"}, "distractor_formula_ids": ["CMF038", "CMF039"], "effective_parameters": {"PDT": 45.1}, "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_0000", "is_gold": true, "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: 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: 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: 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": 50.1}, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=50.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 45.1. 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=50.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 45.1. 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: 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: 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 dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=50.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 45.1. 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=50.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 45.1. 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=50.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 45.1. 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=50.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 45.1. 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=50.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 45.1. 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=45.1 physiological days.\nThe computed value is 0.91 fraction.\nTherefore, the final numeric answer is \\boxed{0.91}.\n#### 0.91", "split": "gold", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF038"}
+{"answer": "1.078", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF009", "CMF014"], "effective_parameters": {"CO2": 478.0, "Coef_B_CO2": 0.23}, "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_0000", "is_gold": true, "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 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 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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 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": 478.0, "Coef_B_CO2": 0.23}, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=478.0 ppm, Coef_B_CO2=0.23 dimensionless. What is the carbon-nitrogen cycling co2 fertilization 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_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=478.0 ppm, Coef_B_CO2=0.23 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 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 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 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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 CO2 fertilization factor. The measured inputs are CO2=478.0 ppm, Coef_B_CO2=0.23 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_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=478.0 ppm, Coef_B_CO2=0.23 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=478.0 ppm, Coef_B_CO2=0.23 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=478.0 ppm, Coef_B_CO2=0.23 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 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 photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=478.0 ppm, Coef_B_CO2=0.23 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final 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=478.0 ppm, Coef_B_CO2=0.23 dimensionless.\nThe computed value is 1.078 fraction.\nTherefore, the final numeric answer is \\boxed{1.078}.\n#### 1.078", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF009"}
+{"answer": "0.62", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF027", "CMF043"], "effective_parameters": {"B_above": 14231.0, "PDT": 27.9, "Y_grain": 4732.0}, "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_t1_0000", "is_gold": true, "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: 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: 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: 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: 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": "Residual with piecewise components", "mode": "chain_2", "parameters": {"B_above": 14231.0, "PDT": 27.9, "Y_grain": 4732.0}, "problem": "Given PDT=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha, compute Stem partitioning index in dry matter partitioning. 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: 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=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha, compute Stem partitioning index in dry matter partitioning. 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: 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: 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: 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=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha, compute Stem partitioning index in dry matter partitioning. 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: 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=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha, compute Stem partitioning index in dry matter partitioning. 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: 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=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha, compute Stem partitioning index in dry matter partitioning. 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: 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=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha, compute Stem partitioning index in dry matter partitioning. 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: 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=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha, compute Stem partitioning index in dry matter partitioning. 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 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=27.9 physiological days, Y_grain=4732.0 kg/ha, B_above=14231.0 kg/ha.\nThe computed value is 0.62 fraction.\nTherefore, the final numeric answer is \\boxed{0.62}.\n#### 0.62", "split": "gold", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF027"}
+{"answer": "0.009", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF019", "CMF023"], "effective_parameters": {"P": 7.803, "Wr": 2016.9, "t_soil": 25.5}, "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_t0_0000", "is_gold": true, "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: 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: 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: 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": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "parameters": {"P": 7.803, "Wr": 2016.9, "t_soil": 25.5}, "problem": "A methane module estimates Methane ebullition emission using intermediate drivers. Use the provided inputs in sequence. The measurements are P=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha. Compute the methane ebullition emission 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: 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: A methane module estimates Methane ebullition emission using intermediate drivers. Use the provided inputs in sequence. The measurements are P=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha. Compute the methane ebullition emission 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 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: 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 module estimates Methane ebullition emission using intermediate drivers. Use the provided inputs in sequence. The measurements are P=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha. Compute the methane ebullition emission 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 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 module estimates Methane ebullition emission using intermediate drivers. Use the provided inputs in sequence. The measurements are P=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha. Compute the methane ebullition emission 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 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: A methane module estimates Methane ebullition emission using intermediate drivers. Use the provided inputs in sequence. The measurements are P=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha. Compute the methane ebullition emission 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 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: A methane module estimates Methane ebullition emission using intermediate drivers. Use the provided inputs in sequence. The measurements are P=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha. Compute the methane ebullition emission 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: 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 module estimates Methane ebullition emission using intermediate drivers. Use the provided inputs in sequence. The measurements are P=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha. Compute the methane ebullition emission 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 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=7.803 g CH4/m2/day, t_soil=25.5 C, Wr=2016.9 kg/ha.\nThe computed value is 0.009 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.009}.\n#### 0.009", "split": "gold", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF019"}
+{"answer": "0.343", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.361, "old_value": 0.314, "parameter": "CH4RiceEfC"}, "distractor_formula_ids": ["CMF021", "CMF016"], "effective_parameters": {"CH4RiceEfC": 0.361, "W": 1583.1, "Wmax": 8400.7}, "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_sensitivity_t0_0000", "is_gold": true, "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: 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: 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: 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 attenuation", "mode": "sensitivity", "parameters": {"CH4RiceEfC": 0.314, "W": 1583.1, "Wmax": 8400.7}, "problem": "A methane calculation involves sensitivity analysis. The current measurements are CH4RiceEfC=0.314 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha. A field technician asks what the plant methane transport factor would become if CH4RiceEfC changed to 0.361. 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: 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 involves sensitivity analysis. The current measurements are CH4RiceEfC=0.314 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha. A field technician asks what the plant methane transport factor would become if CH4RiceEfC changed to 0.361. 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: 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: 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: 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 CH4RiceEfC=0.314 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha. A field technician asks what the plant methane transport factor would become if CH4RiceEfC changed to 0.361. 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: 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 CH4RiceEfC=0.314 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha. A field technician asks what the plant methane transport factor would become if CH4RiceEfC changed to 0.361. 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: 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 involves sensitivity analysis. The current measurements are CH4RiceEfC=0.314 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha. A field technician asks what the plant methane transport factor would become if CH4RiceEfC changed to 0.361. 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: 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 involves sensitivity analysis. The current measurements are CH4RiceEfC=0.314 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha. A field technician asks what the plant methane transport factor would become if CH4RiceEfC changed to 0.361. 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 CH4RiceEfC=0.314 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha. A field technician asks what the plant methane transport factor would become if CH4RiceEfC changed to 0.361. Report the new 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=0.361 fraction, W=1583.1 kg/ha, Wmax=8400.7 kg/ha.\nThe computed value is 0.343 fraction.\nTherefore, the final numeric answer is \\boxed{0.343}.\n#### 0.343", "split": "gold", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF021"}
+{"answer": "19.42", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.92, "old_value": 0.84, "parameter": "NNI"}, "distractor_formula_ids": ["CMF028", "CMF051"], "effective_parameters": {"Am": 28.6, "FCO2": 1.32, "FPA": 0.96, "FT": 0.94, "NNI": 0.92, "WDF": 0.57}, "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_0000", "is_gold": true, "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: 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: 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": "Multiplicative with minimum stress", "mode": "sensitivity", "parameters": {"Am": 28.6, "FCO2": 1.32, "FPA": 0.96, "FT": 0.94, "NNI": 0.84, "WDF": 0.57}, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.84 fraction, WDF=0.57 fraction. A field technician asks what the actual maximum assimilation rate would become if NNI changed to 0.92. 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=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.84 fraction, WDF=0.57 fraction. A field technician asks what the actual maximum assimilation rate would become if NNI changed to 0.92. 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: 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: 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 Am=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.84 fraction, WDF=0.57 fraction. A field technician asks what the actual maximum assimilation rate would become if NNI changed to 0.92. 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=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.84 fraction, WDF=0.57 fraction. A field technician asks what the actual maximum assimilation rate would become if NNI changed to 0.92. 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=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.84 fraction, WDF=0.57 fraction. A field technician asks what the actual maximum assimilation rate would become if NNI changed to 0.92. 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=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.84 fraction, WDF=0.57 fraction. A field technician asks what the actual maximum assimilation rate would become if NNI changed to 0.92. 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: 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 Am=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.84 fraction, WDF=0.57 fraction. A field technician asks what the actual maximum assimilation rate would become if NNI changed to 0.92. 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=28.6 kg CO2/ha/h, FCO2=1.32 fraction, FPA=0.96 fraction, FT=0.94 fraction, NNI=0.92 fraction, WDF=0.57 fraction.\nThe computed value is 19.42 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{19.42}.\n#### 19.42", "split": "gold", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF028"}
+{"answer": "0.739", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF001", "CMF002"], "effective_parameters": {"BD": 1.45, "Saturate_Water": 0.48, "Soil_Water": 0.187}, "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��解水分因子", "name_en": "carbon-nitrogen cycling soil carbon water factor", "precision": 3, "unit": "fraction"}, "id": "CMF012_branch_t1_0000", "is_gold": true, "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 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 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 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": 4, "math_type": "Piecewise exponential", "mode": "branch", "parameters": {"BD": 1.45, "Saturate_Water": 0.48, "Soil_Water": 0.187}, "problem": "For carbon-nitrogen cycling soil carbon, compute the carbon-nitrogen cycling soil carbon water factor. The inputs are Soil_Water=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/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_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: For carbon-nitrogen cycling soil carbon, compute the carbon-nitrogen cycling soil carbon water factor. The inputs are Soil_Water=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/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:", "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 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���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: For carbon-nitrogen cycling soil carbon, compute the carbon-nitrogen cycling soil carbon water factor. The inputs are Soil_Water=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/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:", "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: For carbon-nitrogen cycling soil carbon, compute the carbon-nitrogen cycling soil carbon water factor. The inputs are Soil_Water=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/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:", "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: For carbon-nitrogen cycling soil carbon, compute the carbon-nitrogen cycling soil carbon water factor. The inputs are Soil_Water=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/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:", "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: For carbon-nitrogen cycling soil carbon, compute the carbon-nitrogen cycling soil carbon water factor. The inputs are Soil_Water=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/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:", "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: For carbon-nitrogen cycling soil carbon, compute the carbon-nitrogen cycling soil carbon water factor. The inputs are Soil_Water=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/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 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=0.187 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.45 g/cm3.\nThe computed value is 0.739 fraction.\nTherefore, the final numeric answer is \\boxed{0.739}.\n#### 0.739", "split": "gold", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "wrong_formula_id": "CMF001"}
+{"answer": "47.09", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF008", "CMF006"], "effective_parameters": {"GrainYield": 8458.0, "NUptake": 179.6}, "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_t2_0001", "is_gold": true, "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 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: 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 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": 1, "math_type": "Ratio", "mode": "direct", "parameters": {"GrainYield": 8458.0, "NUptake": 179.6}, "problem": "In a nitrogen, the Nitrogen use efficiency is computed as part of nitrogen modeling. With GrainYield=8458.0 kg/ha, NUptake=179.6 kg N/ha, determine 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 a nitrogen, the Nitrogen use efficiency is computed as part of nitrogen modeling. With GrainYield=8458.0 kg/ha, NUptake=179.6 kg N/ha, determine 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 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 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: In a nitrogen, the Nitrogen use efficiency is computed as part of nitrogen modeling. With GrainYield=8458.0 kg/ha, NUptake=179.6 kg N/ha, determine 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 a nitrogen, the Nitrogen use efficiency is computed as part of nitrogen modeling. With GrainYield=8458.0 kg/ha, NUptake=179.6 kg N/ha, determine 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 a nitrogen, the Nitrogen use efficiency is computed as part of nitrogen modeling. With GrainYield=8458.0 kg/ha, NUptake=179.6 kg N/ha, determine 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 a nitrogen, the Nitrogen use efficiency is computed as part of nitrogen modeling. With GrainYield=8458.0 kg/ha, NUptake=179.6 kg N/ha, determine 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 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 nitrogen, the Nitrogen use efficiency is computed as part of nitrogen modeling. With GrainYield=8458.0 kg/ha, NUptake=179.6 kg N/ha, determine 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=8458.0 kg/ha, NUptake=179.6 kg N/ha.\nThe computed value is 47.09 kg grain/kg N.\nTherefore, the final numeric answer is \\boxed{47.09}.\n#### 47.09", "split": "gold", "subject": "Nitrogen", "template_id": 2, "wrong_formula_id": "CMF008"}
+{"answer": "0.91", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF033", "CMF041"], "effective_parameters": {"PDT": 34.2, "Qt": 16.8, "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_t0_0001", "is_gold": true, "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: 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.", "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: 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": 4, "math_type": "Piecewise power penalty", "mode": "branch", "parameters": {"PDT": 34.2, "Qt": 16.8, "Tmean": 20.6}, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day. Determine the low-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: 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: 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: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day. Determine the low-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: 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: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day. Determine the low-temperature fertility factor and 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=34.2 physiological days, Tmean=20.6 C, Qt=16.8 C day.\nThe computed value is 0.91 fraction.\nTherefore, the final numeric answer is \\boxed{0.91}.\n#### 0.91", "split": "gold", "subject": "Temperature stress", "template_id": 0, "wrong_formula_id": "CMF033"}
+{"answer": "0.98", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF034", "CMF027"], "effective_parameters": {"LAI": 5.14, "k": 0.74}, "family": "growth_yield", "formula": {"expression": "1 - exp(-k * LAI)", "id": "CMF031", "latex": "fPAR = 1 - e^{-k \\cdot LAI}", "name": "光能截获率", "name_en": "Light interception fraction", "precision": 2, "unit": "fraction"}, "id": "CMF031_direct_t2_0001", "is_gold": true, "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 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: 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: 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_formula": "Formula 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]", "knowledge_formula_expr": "Formula: 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", "knowledge_name": "Relevant formula name: Light interception fraction (光能截获率).\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", "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 saturation", "mode": "direct", "parameters": {"LAI": 5.14, "k": 0.74}, "problem": "In a photosynthesis, the Light interception fraction is computed as part of photosynthesis modeling. With k=0.74 dimensionless, LAI=5.14 m2/m2, 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: 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\n\nQ: In a photosynthesis, the Light interception fraction is computed as part of photosynthesis modeling. With k=0.74 dimensionless, LAI=5.14 m2/m2, 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: 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 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: 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 photosynthesis, the Light interception fraction is computed as part of photosynthesis modeling. With k=0.74 dimensionless, LAI=5.14 m2/m2, 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: 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 photosynthesis, the Light interception fraction is computed as part of photosynthesis modeling. With k=0.74 dimensionless, LAI=5.14 m2/m2, 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: 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]\n\nQ: In a photosynthesis, the Light interception fraction is computed as part of photosynthesis modeling. With k=0.74 dimensionless, LAI=5.14 m2/m2, 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: Light interception fraction (光能截获率).\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\n\nQ: In a photosynthesis, the Light interception fraction is computed as part of photosynthesis modeling. With k=0.74 dimensionless, LAI=5.14 m2/m2, 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: 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 a photosynthesis, the Light interception fraction is computed as part of photosynthesis modeling. With k=0.74 dimensionless, LAI=5.14 m2/m2, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:"}, "solution": "Identify the target quantity as Light interception fraction.\nUse the verified formula:\n\\[fPAR = 1 - e^{-k \\cdot LAI}\\]\nIn calculation-friendly notation, this is:\nfPAR = 1 - exp(-k * LAI).\nSubstitute the relevant values:\nk=0.74 dimensionless, LAI=5.14 m2/m2.\nThe computed value is 0.98 fraction.\nTherefore, the final numeric answer is \\boxed{0.98}.\n#### 0.98", "split": "gold", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF034"}
+{"answer": "3.441", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF015", "CMF018"], "effective_parameters": {"OMN": 240.5, "SI": 0.81, "TI": 1.392, "WI": 0.47}, "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_direct_t2_0001", "is_gold": true, "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: 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: 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: 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: 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": "Multiplicative", "mode": "direct", "parameters": {"OMN": 240.5, "SI": 0.81, "TI": 1.392, "WI": 0.47}, "problem": "In a methane, the Fresh organic matter carbon decomposition is computed as part of methane modeling. With WI=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 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: 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 a methane, the Fresh organic matter carbon decomposition is computed as part of methane modeling. With WI=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 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: 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: 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: 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 Fresh organic matter carbon decomposition is computed as part of methane modeling. With WI=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 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: 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 Fresh organic matter carbon decomposition is computed as part of methane modeling. With WI=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 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: 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 a methane, the Fresh organic matter carbon decomposition is computed as part of methane modeling. With WI=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 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: 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 a methane, the Fresh organic matter carbon decomposition is computed as part of methane modeling. With WI=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 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: 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 Fresh organic matter carbon decomposition is computed as part of methane modeling. With WI=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 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 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=0.47 fraction, SI=0.81 fraction, TI=1.392 fraction, OMN=240.5 g C/m2.\nThe computed value is 3.441 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{3.441}.\n#### 3.441", "split": "gold", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF015"}
+{"answer": "5.200", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF020", "CMF025"], "effective_parameters": {"CH4RiceEfC": 0.471, "Ebl": 0.317, "P": 10.855, "W": 2174.7, "Wmax": 12944.4}, "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_0001", "is_gold": true, "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: 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: 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: 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: 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": 4, "math_type": "Chained sum", "mode": "chain_2", "parameters": {"CH4RiceEfC": 0.471, "Ebl": 0.317, "P": 10.855, "W": 2174.7, "Wmax": 12944.4}, "problem": "Given Ebl=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 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=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 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: 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: 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 Ebl=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 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=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 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=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 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=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 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: 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 Ebl=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 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=0.317 g CH4/m2/day, P=10.855 g CH4/m2/day, CH4RiceEfC=0.471 fraction, W=2174.7 kg/ha, Wmax=12944.4 kg/ha.\nThe computed value is 5.200 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{5.200}.\n#### 5.200", "split": "gold", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF020"}
+{"answer": "0.0061", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF042", "CMF035"], "effective_parameters": {"NNI": 0.72, "RP": 0.0085, "WDF": 0.95}, "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_t2_0000", "is_gold": true, "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: 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: 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: 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": "Minimum stress multiplier", "mode": "chain_2", "parameters": {"NNI": 0.72, "RP": 0.0085, "WDF": 0.95}, "problem": "In leaf area growth, the Actual LAI relative growth rate is obtained by chaining intermediate calculations. With inputs RP=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 fraction, 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: 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, the Actual LAI relative growth rate is obtained by chaining intermediate calculations. With inputs RP=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 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: 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: 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 leaf area growth, the Actual LAI relative growth rate is obtained by chaining intermediate calculations. With inputs RP=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 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 leaf area growth, the Actual LAI relative growth rate is obtained by chaining intermediate calculations. With inputs RP=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 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 leaf area growth, the Actual LAI relative growth rate is obtained by chaining intermediate calculations. With inputs RP=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 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 leaf area growth, the Actual LAI relative growth rate is obtained by chaining intermediate calculations. With inputs RP=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 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: 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 leaf area growth, the Actual LAI relative growth rate is obtained by chaining intermediate calculations. With inputs RP=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 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=0.0085 d-1, NNI=0.72 fraction, WDF=0.95 fraction.\nThe computed value is 0.0061 d-1.\nTherefore, the final numeric answer is \\boxed{0.0061}.\n#### 0.0061", "split": "gold", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF042"}
+{"answer": "1.40", "benchmark_version": "cropmath-v1", "change": {"new_value": 409.2, "old_value": 355.8, "parameter": "GDD"}, "distractor_formula_ids": ["CMF034", "CMF054"], "effective_parameters": {"GDD": 409.2, "LAI0": 0.241, "RP": 0.0043}, "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_sensitivity_t1_0001", "is_gold": true, "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: 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: 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": 2, "math_type": "Exponential growth", "mode": "sensitivity", "parameters": {"GDD": 355.8, "LAI0": 0.241, "RP": 0.0043}, "problem": "Given current measurements LAI0=0.241 m2/m2, RP=0.0043 d-1, GDD=355.8 C day in leaf area growth, what would the Potential leaf area index be if GDD were adjusted to 409.2? Provide 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: 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 current measurements LAI0=0.241 m2/m2, RP=0.0043 d-1, GDD=355.8 C day in leaf area growth, what would the Potential leaf area index be if GDD were adjusted to 409.2? Provide 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: 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: 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 LAI0=0.241 m2/m2, RP=0.0043 d-1, GDD=355.8 C day in leaf area growth, what would the Potential leaf area index be if GDD were adjusted to 409.2? Provide 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: 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 LAI0=0.241 m2/m2, RP=0.0043 d-1, GDD=355.8 C day in leaf area growth, what would the Potential leaf area index be if GDD were adjusted to 409.2? Provide 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: 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 current measurements LAI0=0.241 m2/m2, RP=0.0043 d-1, GDD=355.8 C day in leaf area growth, what would the Potential leaf area index be if GDD were adjusted to 409.2? Provide 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: 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 current measurements LAI0=0.241 m2/m2, RP=0.0043 d-1, GDD=355.8 C day in leaf area growth, what would the Potential leaf area index be if GDD were adjusted to 409.2? Provide 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: 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 LAI0=0.241 m2/m2, RP=0.0043 d-1, GDD=355.8 C day in leaf area growth, what would the Potential leaf area index be if GDD were adjusted to 409.2? Provide the 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=0.241 m2/m2, RP=0.0043 d-1, GDD=409.2 C day.\nThe computed value is 1.40 m2/m2.\nTherefore, the final numeric answer is \\boxed{1.40}.\n#### 1.40", "split": "gold", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF034"}
+{"answer": "0.49", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF030", "CMF033"], "effective_parameters": {"B_above": 10593.0, "PDT": 56.9, "Y_grain": 4544.0}, "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_t2_0000", "is_gold": true, "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: 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: 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: 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: 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": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "parameters": {"B_above": 10593.0, "PDT": 56.9, "Y_grain": 4544.0}, "problem": "Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days. 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: 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: 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: 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 Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days. 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: 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 Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days. 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 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=4544.0 kg/ha, B_above=10593.0 kg/ha, PDT=56.9 physiological days.\nThe computed value is 0.49 fraction.\nTherefore, the final numeric answer is \\boxed{0.49}.\n#### 0.49", "split": "gold", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF030"}
+{"answer": "0.074", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF016", "CMF021"], "effective_parameters": {"Eh": 80.3}, "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_t0_0000", "is_gold": true, "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: 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: 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": 2, "math_type": "Piecewise exponential", "mode": "direct", "parameters": {"Eh": 80.3}, "problem": "A methane calculation estimates Redox potential factor. The measured inputs are Eh=80.3 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=80.3 mV. What is the redox potential 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: 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: 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 Redox potential factor. The measured inputs are Eh=80.3 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=80.3 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=80.3 mV. What is the redox potential 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: 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 estimates Redox potential factor. The measured inputs are Eh=80.3 mV. What is the redox potential 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: 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 Redox potential factor. The measured inputs are Eh=80.3 mV. What is the redox potential factor? Give 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=80.3 mV.\nThe computed value is 0.074 fraction.\nTherefore, the final numeric answer is \\boxed{0.074}.\n#### 0.074", "split": "gold", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF016"}
+{"answer": "0.14", "benchmark_version": "cropmath-v1", "change": {"new_value": 48.1, "old_value": 41.8, "parameter": "PDT"}, "distractor_formula_ids": ["CMF043", "CMF046"], "effective_parameters": {"PDT": 48.1}, "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_0001", "is_gold": true, "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: 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: 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: 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: 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": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "parameters": {"PDT": 41.8}, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=41.8 physiological days. After changing PDT from 41.8 to 48.1, 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=41.8 physiological days. After changing PDT from 41.8 to 48.1, 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: 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: 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: 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=41.8 physiological days. After changing PDT from 41.8 to 48.1, 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=41.8 physiological days. After changing PDT from 41.8 to 48.1, 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=41.8 physiological days. After changing PDT from 41.8 to 48.1, 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=41.8 physiological days. After changing PDT from 41.8 to 48.1, 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: 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 Green leaf partitioning index responds to parameter changes. Baseline: PDT=41.8 physiological days. After changing PDT from 41.8 to 48.1, 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=48.1 physiological days.\nThe computed value is 0.14 fraction.\nTherefore, the final numeric answer is \\boxed{0.14}.\n#### 0.14", "split": "gold", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF043"}
+{"answer": "22.903", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF007", "CMF004"], "effective_parameters": {"Day_Length_hour": 11.54, "F_CO2": 1.24, "F_T": 0.76, "F_Water": 0.91, "Photo_Rate_opt": 53.57}, "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_0000", "is_gold": true, "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 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 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 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": "Multiplicative conversion", "mode": "direct", "parameters": {"Day_Length_hour": 11.54, "F_CO2": 1.24, "F_T": 0.76, "F_Water": 0.91, "Photo_Rate_opt": 53.57}, "problem": "Given Photo_Rate_opt=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 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=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 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 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 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 Photo_Rate_opt=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 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=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 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=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 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=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 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 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: Given Photo_Rate_opt=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 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=53.57 umol CO2/m2/s, F_T=0.76 fraction, F_Water=0.91 fraction, F_CO2=1.24 fraction, Day_Length_hour=11.54 h.\nThe computed value is 22.903 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{22.903}.\n#### 22.903", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "wrong_formula_id": "CMF007"}
+{"answer": "0.0035", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF038", "CMF035"], "effective_parameters": {"NNI": 1.02, "RP": 0.0047, "WDF": 0.75}, "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_0001", "is_gold": true, "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: 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: 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: 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: 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": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "parameters": {"NNI": 1.02, "RP": 0.0047, "WDF": 0.75}, "problem": "Given RP=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 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=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 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: 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: 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: 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 RP=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 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=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 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=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 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=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 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: 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 RP=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 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=0.0047 d-1, NNI=1.02 fraction, WDF=0.75 fraction.\nThe computed value is 0.0035 d-1.\nTherefore, the final numeric answer is \\boxed{0.0035}.\n#### 0.0035", "split": "gold", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF038"}
+{"answer": "1.630", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF055", "CMF061"], "effective_parameters": {"CO2": 804.0, "CO2_RUE": 0.18}, "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_t0_0001", "is_gold": true, "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 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 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 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": 4, "math_type": "Piecewise linear cap", "mode": "branch", "parameters": {"CO2": 804.0, "CO2_RUE": 0.18}, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are CO2=804.0 ppm, CO2_RUE=0.18 fraction. Determine the simple co2 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_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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are CO2=804.0 ppm, CO2_RUE=0.18 fraction. Determine the simple co2 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 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 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 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are CO2=804.0 ppm, CO2_RUE=0.18 fraction. Determine the simple co2 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 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are CO2=804.0 ppm, CO2_RUE=0.18 fraction. Determine the simple co2 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 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are CO2=804.0 ppm, CO2_RUE=0.18 fraction. Determine the simple co2 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 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are CO2=804.0 ppm, CO2_RUE=0.18 fraction. Determine the simple co2 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 CO2=804.0 ppm, CO2_RUE=0.18 fraction. Determine the simple co2 response and 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=804.0 ppm, CO2_RUE=0.18 fraction.\nThe computed value is 1.630 fraction.\nTherefore, the final numeric answer is \\boxed{1.630}.\n#### 1.630", "split": "gold", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF055"}
+{"answer": "0.656", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF012", "CMF013"], "effective_parameters": {"Coef_Plant_TN_uptake": 0.108, "DVI": 1.55, "NPP": 16.83, "N_requirement_max_day": 0.68, "N_shoot": 3.2, "Ratio_N_C_photo_product": 0.039}, "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_chain_2_t1_0001", "is_gold": true, "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 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 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 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 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": "Piecewise minimum", "mode": "chain_2", "parameters": {"Coef_Plant_TN_uptake": 0.108, "DVI": 1.55, "NPP": 16.83, "N_requirement_max_day": 0.68, "N_shoot": 3.2, "Ratio_N_C_photo_product": 0.039}, "problem": "Given N_shoot=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless, compute carbon-nitrogen cycling crop nitrogen demand in carbon-nitrogen cycling nitrogen. This requires sequential calculation through intermediate variables. 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: Given N_shoot=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless, compute carbon-nitrogen cycling crop nitrogen demand in carbon-nitrogen cycling nitrogen. This requires sequential calculation through intermediate variables. 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 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 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 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 N_shoot=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless, compute carbon-nitrogen cycling crop nitrogen demand in carbon-nitrogen cycling nitrogen. This requires sequential calculation through intermediate variables. 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: Given N_shoot=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless, compute carbon-nitrogen cycling crop nitrogen demand in carbon-nitrogen cycling nitrogen. This requires sequential calculation through intermediate variables. 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: Given N_shoot=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless, compute carbon-nitrogen cycling crop nitrogen demand in carbon-nitrogen cycling nitrogen. This requires sequential calculation through intermediate variables. 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: Given N_shoot=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless, compute carbon-nitrogen cycling crop nitrogen demand in carbon-nitrogen cycling nitrogen. This requires sequential calculation through intermediate variables. 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 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 N_shoot=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless, compute carbon-nitrogen cycling crop nitrogen demand in carbon-nitrogen cycling nitrogen. This requires sequential calculation through intermediate variables. 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=3.2 g N/m2, Coef_Plant_TN_uptake=0.108 d-1, N_requirement_max_day=0.68 g N/m2/day, NPP=16.83 g C/m2/day, Ratio_N_C_photo_product=0.039 g N/g C, DVI=1.55 dimensionless.\nThe computed value is 0.656 g N/m2/day.\nTherefore, the final numeric answer is \\boxed{0.656}.\n#### 0.656", "split": "gold", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "wrong_formula_id": "CMF012"}
+{"answer": "1.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF043", "CMF044"], "effective_parameters": {"PDT": 25.1, "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_t2_0001", "is_gold": true, "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: 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: 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: 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: 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": "Piecewise exponential decay", "mode": "direct", "parameters": {"PDT": 25.1, "a": 0.053}, "problem": "In a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=25.1 physiological days, a=0.053 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=25.1 physiological days, a=0.053 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: 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: 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=25.1 physiological days, a=0.053 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=25.1 physiological days, a=0.053 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=25.1 physiological days, a=0.053 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=25.1 physiological days, a=0.053 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=25.1 physiological days, a=0.053 dimensionless, determine 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=25.1 physiological days, a=0.053 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "gold", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF043"}
+{"answer": "4924.7", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF060", "CMF061"], "effective_parameters": {"Biomass": 8346.9, "HI": 0.59}, "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_0001", "is_gold": true, "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 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 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 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.", "level": 1, "math_type": "Product", "mode": "direct", "parameters": {"Biomass": 8346.9, "HI": 0.59}, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=8346.9 kg/ha, HI=0.59 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=8346.9 kg/ha, HI=0.59 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 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 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: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=8346.9 kg/ha, HI=0.59 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=8346.9 kg/ha, HI=0.59 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=8346.9 kg/ha, HI=0.59 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=8346.9 kg/ha, HI=0.59 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 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 yield is computed as part of environmental-response formula calculation. With Biomass=8346.9 kg/ha, HI=0.59 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=8346.9 kg/ha, HI=0.59 fraction.\nThe computed value is 4924.7 kg/ha.\nTherefore, the final numeric answer is \\boxed{4924.7}.\n#### 4924.7", "split": "gold", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF060"}
+{"answer": "442.59", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF013", "CMF014"], "effective_parameters": {"Coef_K": 0.66, "Coef_rou": 0.08, "LAI": 3.6, "Radi": 622.6}, "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_direct_t2_0001", "is_gold": true, "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 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\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 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 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": "Exponential ratio", "mode": "direct", "parameters": {"Coef_K": 0.66, "Coef_rou": 0.08, "LAI": 3.6, "Radi": 622.6}, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Radi=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 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 a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Radi=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 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: 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\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 canopy absorbed PAR is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Radi=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 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 a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Radi=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 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 a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Radi=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 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 a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Radi=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 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: 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 canopy absorbed PAR is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Radi=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 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=622.6 W/m2, LAI=3.6 m2/m2, Coef_rou=0.08 fraction, Coef_K=0.66 dimensionless.\nThe computed value is 442.59 umol/m2/s.\nTherefore, the final numeric answer is \\boxed{442.59}.\n#### 442.59", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF013"}
+{"answer": "0.493", "benchmark_version": "cropmath-v1", "change": {"new_value": 31.1, "old_value": 34.6, "parameter": "PDT"}, "distractor_formula_ids": ["CMF034", "CMF047"], "effective_parameters": {"PDT": 31.1}, "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_t1_0000", "is_gold": true, "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: 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: 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: 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: 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": "Linear", "mode": "sensitivity", "parameters": {"PDT": 34.6}, "problem": "Given current measurements PDT=34.6 physiological days in photosynthesis, what would the Canopy extinction coefficient be if PDT were adjusted to 31.1? Provide 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: Given current measurements PDT=34.6 physiological days in photosynthesis, what would the Canopy extinction coefficient be if PDT were adjusted to 31.1? Provide 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: 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: 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: 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 PDT=34.6 physiological days in photosynthesis, what would the Canopy extinction coefficient be if PDT were adjusted to 31.1? Provide 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: Given current measurements PDT=34.6 physiological days in photosynthesis, what would the Canopy extinction coefficient be if PDT were adjusted to 31.1? Provide 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: Given current measurements PDT=34.6 physiological days in photosynthesis, what would the Canopy extinction coefficient be if PDT were adjusted to 31.1? Provide 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: Given current measurements PDT=34.6 physiological days in photosynthesis, what would the Canopy extinction coefficient be if PDT were adjusted to 31.1? Provide 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: 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 PDT=34.6 physiological days in photosynthesis, what would the Canopy extinction coefficient be if PDT were adjusted to 31.1? Provide 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=31.1 physiological days.\nThe computed value is 0.493 dimensionless.\nTherefore, the final numeric answer is \\boxed{0.493}.\n#### 0.493", "split": "gold", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF034"}
+{"answer": "0.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 15.39, "old_value": 13.38, "parameter": "P"}, "distractor_formula_ids": ["CMF032", "CMF041"], "effective_parameters": {"P": 15.39, "PS": 0.322, "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_sensitivity_t0_0000", "is_gold": true, "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<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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: 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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 3, "math_type": "Piecewise quadratic", "mode": "sensitivity", "parameters": {"P": 13.38, "PS": 0.322, "Po": 10.5}, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are P=13.38 h, Po=10.5 h, PS=0.322 dimensionless. A field technician asks what the relative photoperiod effectiveness would become if P changed to 15.39. 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: 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: A phenology calculation involves sensitivity analysis. The current measurements are P=13.38 h, Po=10.5 h, PS=0.322 dimensionless. A field technician asks what the relative photoperiod effectiveness would become if P changed to 15.39. 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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: 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: A phenology calculation involves sensitivity analysis. The current measurements are P=13.38 h, Po=10.5 h, PS=0.322 dimensionless. A field technician asks what the relative photoperiod effectiveness would become if P changed to 15.39. 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 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: A phenology calculation involves sensitivity analysis. The current measurements are P=13.38 h, Po=10.5 h, PS=0.322 dimensionless. A field technician asks what the relative photoperiod effectiveness would become if P changed to 15.39. 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 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: A phenology calculation involves sensitivity analysis. The current measurements are P=13.38 h, Po=10.5 h, PS=0.322 dimensionless. A field technician asks what the relative photoperiod effectiveness would become if P changed to 15.39. 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 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: A phenology calculation involves sensitivity analysis. The current measurements are P=13.38 h, Po=10.5 h, PS=0.322 dimensionless. A field technician asks what the relative photoperiod effectiveness would become if P changed to 15.39. 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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 P=13.38 h, Po=10.5 h, PS=0.322 dimensionless. A field technician asks what the relative photoperiod effectiveness would become if P changed to 15.39. Report the new 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=15.39 h, Po=10.5 h, PS=0.322 dimensionless.\nThe computed value is 0.00 fraction.\nTherefore, the final numeric answer is \\boxed{0.00}.\n#### 0.00", "split": "gold", "subject": "Phenology", "template_id": 0, "wrong_formula_id": "CMF032"}
+{"answer": "1.43", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF035", "CMF052"], "effective_parameters": {"Cx": 609.0, "beta": 0.73}, "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_t0_0001", "is_gold": true, "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 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: 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: 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": "Logarithmic", "mode": "direct", "parameters": {"Cx": 609.0, "beta": 0.73}, "problem": "A photosynthesis calculation estimates CO2 response factor. The measured inputs are Cx=609.0 ppm, beta=0.73 dimensionless. What is the co2 response 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: 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 estimates CO2 response factor. The measured inputs are Cx=609.0 ppm, beta=0.73 dimensionless. What is the co2 response 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: 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 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 photosynthesis calculation estimates CO2 response factor. The measured inputs are Cx=609.0 ppm, beta=0.73 dimensionless. What is the co2 response 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: 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 estimates CO2 response factor. The measured inputs are Cx=609.0 ppm, beta=0.73 dimensionless. What is the co2 response 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: 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 estimates CO2 response factor. The measured inputs are Cx=609.0 ppm, beta=0.73 dimensionless. What is the co2 response 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: 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 estimates CO2 response factor. The measured inputs are Cx=609.0 ppm, beta=0.73 dimensionless. What is the co2 response 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: 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 photosynthesis calculation estimates CO2 response factor. The measured inputs are Cx=609.0 ppm, beta=0.73 dimensionless. What is the co2 response factor? Give the 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=609.0 ppm, beta=0.73 dimensionless.\nThe computed value is 1.43 fraction.\nTherefore, the final numeric answer is \\boxed{1.43}.\n#### 1.43", "split": "gold", "subject": "Photosynthesis", "template_id": 0, "wrong_formula_id": "CMF035"}
+{"answer": "0.01", "benchmark_version": "cropmath-v1", "change": {"new_value": 42.0, "old_value": 40.0, "parameter": "Tmax_day"}, "distractor_formula_ids": ["CMF037", "CMF032"], "effective_parameters": {"PDT": 35.8, "Tmax_day": 42.0}, "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_t0_0001", "is_gold": true, "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<=T<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "parameters": {"PDT": 35.8, "Tmax_day": 40.0}, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=35.8 physiological days, Tmax_day=40.0 C. A field technician asks what the high-temperature fertility factor would become if Tmax_day changed to 42.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.8 physiological days, Tmax_day=40.0 C. A field technician asks what the high-temperature fertility factor would become if Tmax_day changed to 42.0. 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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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 calculation involves sensitivity analysis. The current measurements are PDT=35.8 physiological days, Tmax_day=40.0 C. A field technician asks what the high-temperature fertility factor would become if Tmax_day changed to 42.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.8 physiological days, Tmax_day=40.0 C. A field technician asks what the high-temperature fertility factor would become if Tmax_day changed to 42.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.8 physiological days, Tmax_day=40.0 C. A field technician asks what the high-temperature fertility factor would become if Tmax_day changed to 42.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.8 physiological days, Tmax_day=40.0 C. A field technician asks what the high-temperature fertility factor would become if Tmax_day changed to 42.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.8 physiological days, Tmax_day=40.0 C. A field technician asks what the high-temperature fertility factor would become if Tmax_day changed to 42.0. Report the new 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=35.8 physiological days, Tmax_day=42.0 C.\nThe computed value is 0.01 fraction.\nTherefore, the final numeric answer is \\boxed{0.01}.\n#### 0.01", "split": "gold", "subject": "Temperature stress", "template_id": 0, "wrong_formula_id": "CMF037"}
+{"answer": "1.000", "benchmark_version": "cropmath-v1", "change": {"new_value": -216.2, "old_value": -188.0, "parameter": "Eh"}, "distractor_formula_ids": ["CMF024", "CMF022"], "effective_parameters": {"Eh": -216.2}, "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_0001", "is_gold": true, "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 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: 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: 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: 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": "Piecewise exponential", "mode": "sensitivity", "parameters": {"Eh": -188.0}, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-188.0 mV. A field technician asks what the redox potential factor would become if Eh changed to -216.2. 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=-188.0 mV. A field technician asks what the redox potential factor would become if Eh changed to -216.2. 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: 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: 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 involves sensitivity analysis. The current measurements are Eh=-188.0 mV. A field technician asks what the redox potential factor would become if Eh changed to -216.2. 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=-188.0 mV. A field technician asks what the redox potential factor would become if Eh changed to -216.2. 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=-188.0 mV. A field technician asks what the redox potential factor would become if Eh changed to -216.2. 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=-188.0 mV. A field technician asks what the redox potential factor would become if Eh changed to -216.2. 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 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 involves sensitivity analysis. The current measurements are Eh=-188.0 mV. A field technician asks what the redox potential factor would become if Eh changed to -216.2. 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=-216.2 mV.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "gold", "subject": "Methane", "template_id": 0, "wrong_formula_id": "CMF024"}
+{"answer": "8.87", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF053", "CMF049"], "effective_parameters": {"GDD": 599.1, "LAI0": 0.088, "RP": 0.0077}, "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_chain_2_t2_0001", "is_gold": true, "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: 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: 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: 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": 3, "math_type": "Exponential growth", "mode": "chain_2", "parameters": {"GDD": 599.1, "LAI0": 0.088, "RP": 0.0077}, "problem": "In leaf area growth, the Potential leaf area index is obtained by chaining intermediate calculations. With inputs LAI0=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day, determine 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: 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: In leaf area growth, the Potential leaf area index is obtained by chaining intermediate calculations. With inputs LAI0=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day, determine 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: 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: 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 leaf area growth, the Potential leaf area index is obtained by chaining intermediate calculations. With inputs LAI0=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day, determine 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: 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, the Potential leaf area index is obtained by chaining intermediate calculations. With inputs LAI0=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day, determine 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: 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: In leaf area growth, the Potential leaf area index is obtained by chaining intermediate calculations. With inputs LAI0=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day, determine 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: 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: In leaf area growth, the Potential leaf area index is obtained by chaining intermediate calculations. With inputs LAI0=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day, determine 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: 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, the Potential leaf area index is obtained by chaining intermediate calculations. With inputs LAI0=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day, determine the 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=0.088 m2/m2, RP=0.0077 d-1, GDD=599.1 C day.\nThe computed value is 8.87 m2/m2.\nTherefore, the final numeric answer is \\boxed{8.87}.\n#### 8.87", "split": "gold", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF053"}
+{"answer": "0.33", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF035", "CMF047"], "effective_parameters": {"B_above": 9512.0, "PDT": 42.5, "Y_grain": 3749.0}, "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_t2_0001", "is_gold": true, "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: 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: 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: 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: 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": "Multi-step sigmoid", "mode": "branch", "parameters": {"B_above": 9512.0, "PDT": 42.5, "Y_grain": 3749.0}, "problem": "Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days. 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: 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: 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days. 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: 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: Calculate Panicle partitioning index in dry matter partitioning with these inputs: Y_grain=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days. 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 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=3749.0 kg/ha, B_above=9512.0 kg/ha, PDT=42.5 physiological days.\nThe computed value is 0.33 fraction.\nTherefore, the final numeric answer is \\boxed{0.33}.\n#### 0.33", "split": "gold", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF035"}
+{"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 39.0, "old_value": 35.3, "parameter": "PDT"}, "distractor_formula_ids": ["CMF033", "CMF046"], "effective_parameters": {"PDT": 39.0, "Tmax_day": 34.3}, "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_t0_0000", "is_gold": true, "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: 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: 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: 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": "Piecewise logistic", "mode": "sensitivity", "parameters": {"PDT": 35.3, "Tmax_day": 34.3}, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=35.3 physiological days, Tmax_day=34.3 C. A field technician asks what the high-temperature fertility factor would become if PDT changed to 39.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.3 physiological days, Tmax_day=34.3 C. A field technician asks what the high-temperature fertility factor would become if PDT changed to 39.0. 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: 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: 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.3 physiological days, Tmax_day=34.3 C. A field technician asks what the high-temperature fertility factor would become if PDT changed to 39.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.3 physiological days, Tmax_day=34.3 C. A field technician asks what the high-temperature fertility factor would become if PDT changed to 39.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.3 physiological days, Tmax_day=34.3 C. A field technician asks what the high-temperature fertility factor would become if PDT changed to 39.0. 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: 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 calculation involves sensitivity analysis. The current measurements are PDT=35.3 physiological days, Tmax_day=34.3 C. A field technician asks what the high-temperature fertility factor would become if PDT changed to 39.0. 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: 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: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=35.3 physiological days, Tmax_day=34.3 C. A field technician asks what the high-temperature fertility factor would become if PDT changed to 39.0. Report the new 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=39.0 physiological days, Tmax_day=34.3 C.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "gold", "subject": "Temperature stress", "template_id": 0, "wrong_formula_id": "CMF033"}
+{"answer": "4.368", "benchmark_version": "cropmath-v1", "change": {"new_value": 262.2, "old_value": 291.3, "parameter": "Biom_root"}, "distractor_formula_ids": ["CMF010", "CMF014"], "effective_parameters": {"Biom_root": 262.2, "Biom_shoot": 84.7, "Plant_TN_uptake": 17.89}, "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_0000", "is_gold": true, "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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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 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 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 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": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "parameters": {"Biom_root": 291.3, "Biom_shoot": 84.7, "Plant_TN_uptake": 17.89}, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=291.3 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_root changed to 262.2. 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=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=291.3 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_root changed to 262.2. 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 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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 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=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=291.3 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_root changed to 262.2. 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=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=291.3 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_root changed to 262.2. 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=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=291.3 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_root changed to 262.2. 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=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=291.3 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_root changed to 262.2. 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 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 nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=291.3 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_root changed to 262.2. 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=17.89 g N/m2, Biom_shoot=84.7 g C/m2, Biom_root=262.2 g C/m2.\nThe computed value is 4.368 g N/m2.\nTherefore, the final numeric answer is \\boxed{4.368}.\n#### 4.368", "split": "gold", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "wrong_formula_id": "CMF010"}
+{"answer": "46.309", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF001", "CMF005"], "effective_parameters": {"Coef_alfa": 20.78, "Coef_beta": 325.9, "Leaf_N": 2.9, "PAR": 1081.6}, "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_0001", "is_gold": true, "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 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 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 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 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": "Michaelis-Menten response", "mode": "chain_2", "parameters": {"Coef_alfa": 20.78, "Coef_beta": 325.9, "Leaf_N": 2.9, "PAR": 1081.6}, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 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=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 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 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 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 optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 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=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 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=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 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=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 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 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, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 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=20.78 umol CO2/gN/s, Coef_beta=325.9 umol/m2/s, PAR=1081.6 umol/m2/s, Leaf_N=2.9 gN/m2.\nThe computed value is 46.309 umol CO2/m2/s.\nTherefore, the final numeric answer is \\boxed{46.309}.\n#### 46.309", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF001"}
+{"answer": "1.406", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF058", "CMF062"], "effective_parameters": {"CO2": 670.0, "CO2_RUE": 0.127}, "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_0000", "is_gold": true, "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 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 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 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 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": 2, "math_type": "Piecewise linear cap", "mode": "direct", "parameters": {"CO2": 670.0, "CO2_RUE": 0.127}, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=670.0 ppm, CO2_RUE=0.127 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=670.0 ppm, CO2_RUE=0.127 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 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 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 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=670.0 ppm, CO2_RUE=0.127 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=670.0 ppm, CO2_RUE=0.127 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=670.0 ppm, CO2_RUE=0.127 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=670.0 ppm, CO2_RUE=0.127 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 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 CO2 response. The measured inputs are CO2=670.0 ppm, CO2_RUE=0.127 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=670.0 ppm, CO2_RUE=0.127 fraction.\nThe computed value is 1.406 fraction.\nTherefore, the final numeric answer is \\boxed{1.406}.\n#### 1.406", "split": "gold", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF058"}
+{"answer": "0.40", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF035", "CMF042"], "effective_parameters": {"P": 11.92, "PS": 0.299, "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_direct_t1_0000", "is_gold": true, "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: 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: 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: 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: 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": "Piecewise quadratic", "mode": "direct", "parameters": {"P": 11.92, "PS": 0.299, "Po": 10.5}, "problem": "Given P=11.92 h, Po=10.5 h, PS=0.299 dimensionless, calculate the Relative photoperiod 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: 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=11.92 h, Po=10.5 h, PS=0.299 dimensionless, calculate the Relative photoperiod 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: 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: 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: 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 P=11.92 h, Po=10.5 h, PS=0.299 dimensionless, calculate the Relative photoperiod 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 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=11.92 h, Po=10.5 h, PS=0.299 dimensionless, calculate the Relative photoperiod 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 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=11.92 h, Po=10.5 h, PS=0.299 dimensionless, calculate the Relative photoperiod 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 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=11.92 h, Po=10.5 h, PS=0.299 dimensionless, calculate the Relative photoperiod 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: 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 P=11.92 h, Po=10.5 h, PS=0.299 dimensionless, calculate the Relative photoperiod effectiveness for a phenology. The result should be 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=11.92 h, Po=10.5 h, PS=0.299 dimensionless.\nThe computed value is 0.40 fraction.\nTherefore, the final numeric answer is \\boxed{0.40}.\n#### 0.40", "split": "gold", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF035"}
+{"answer": "154.75", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF052", "CMF026"], "effective_parameters": {"GDD": 634.4, "SLAc": 213.4}, "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_direct_t1_0001", "is_gold": true, "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: 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: 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: 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: 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": 2, "math_type": "Piecewise polynomial", "mode": "direct", "parameters": {"GDD": 634.4, "SLAc": 213.4}, "problem": "Given GDD=634.4 C day, SLAc=213.4 cm2/g, calculate the Specific leaf area for a leaf area growth. The result should be 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: Given GDD=634.4 C day, SLAc=213.4 cm2/g, calculate the Specific leaf area for a leaf area growth. The result should be 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: 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: 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 GDD=634.4 C day, SLAc=213.4 cm2/g, calculate the Specific leaf area for a leaf area growth. The result should be 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: Given GDD=634.4 C day, SLAc=213.4 cm2/g, calculate the Specific leaf area for a leaf area growth. The result should be 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: Given GDD=634.4 C day, SLAc=213.4 cm2/g, calculate the Specific leaf area for a leaf area growth. The result should be 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: Given GDD=634.4 C day, SLAc=213.4 cm2/g, calculate the Specific leaf area for a leaf area growth. The result should be 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: 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 GDD=634.4 C day, SLAc=213.4 cm2/g, calculate the Specific leaf area for a leaf area growth. The result should be 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=634.4 C day, SLAc=213.4 cm2/g.\nThe computed value is 154.75 cm2/g.\nTherefore, the final numeric answer is \\boxed{154.75}.\n#### 154.75", "split": "gold", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF052"}
+{"answer": "0.412", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.13, "old_value": 1.03, "parameter": "ARID"}, "distractor_formula_ids": ["CMF062", "CMF055"], "effective_parameters": {"ARID": 1.13, "S_Water": 0.52}, "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_t2_0000", "is_gold": true, "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 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 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": 1.03, "S_Water": 0.52}, "problem": "In environmental-response formula system, we analyze how environmental-response water stress response responds to parameter changes. Baseline: ARID=1.03 dimensionless, S_Water=0.52 dimensionless. After changing ARID from 1.03 to 1.13, compute the new environmental-response water 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_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: In environmental-response formula system, we analyze how environmental-response water stress response responds to parameter changes. Baseline: ARID=1.03 dimensionless, S_Water=0.52 dimensionless. After changing ARID from 1.03 to 1.13, compute the new environmental-response water 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 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 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 environmental-response formula system, we analyze how environmental-response water stress response responds to parameter changes. Baseline: ARID=1.03 dimensionless, S_Water=0.52 dimensionless. After changing ARID from 1.03 to 1.13, compute the new environmental-response water 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 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 environmental-response formula system, we analyze how environmental-response water stress response responds to parameter changes. Baseline: ARID=1.03 dimensionless, S_Water=0.52 dimensionless. After changing ARID from 1.03 to 1.13, compute the new environmental-response water 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 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: In environmental-response formula system, we analyze how environmental-response water stress response responds to parameter changes. Baseline: ARID=1.03 dimensionless, S_Water=0.52 dimensionless. After changing ARID from 1.03 to 1.13, compute the new environmental-response water 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 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: In environmental-response formula system, we analyze how environmental-response water stress response responds to parameter changes. Baseline: ARID=1.03 dimensionless, S_Water=0.52 dimensionless. After changing ARID from 1.03 to 1.13, compute the new environmental-response water 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 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 water stress response responds to parameter changes. Baseline: ARID=1.03 dimensionless, S_Water=0.52 dimensionless. After changing ARID from 1.03 to 1.13, compute the new environmental-response water stress response. Give 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=1.13 dimensionless, S_Water=0.52 dimensionless.\nThe computed value is 0.412 fraction.\nTherefore, the final numeric answer is \\boxed{0.412}.\n#### 0.412", "split": "gold", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF062"}
+{"answer": "0.90", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF031", "CMF046"], "effective_parameters": {"PDT": 33.3, "Qt": 18.6, "Tmean": 18.9}, "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_branch_t0_0000", "is_gold": true, "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: 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: 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: 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: 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": 4, "math_type": "Piecewise power penalty", "mode": "branch", "parameters": {"PDT": 33.3, "Qt": 18.6, "Tmean": 18.9}, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day. Determine the low-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: 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: 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: 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 temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day. Determine the low-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: 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 involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day. Determine the low-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: 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 temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day. Determine the low-temperature fertility factor and 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=33.3 physiological days, Tmean=18.9 C, Qt=18.6 C day.\nThe computed value is 0.90 fraction.\nTherefore, the final numeric answer is \\boxed{0.90}.\n#### 0.90", "split": "gold", "subject": "Temperature stress", "template_id": 0, "wrong_formula_id": "CMF031"}
+{"answer": "1.00", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF045", "CMF038"], "effective_parameters": {"PDT": 23.3, "a": 0.029}, "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_t2_0000", "is_gold": true, "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: 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: 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_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: 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": "Piecewise exponential decay", "mode": "direct", "parameters": {"PDT": 23.3, "a": 0.029}, "problem": "In a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=23.3 physiological days, a=0.029 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=23.3 physiological days, a=0.029 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: 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: 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: 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 Physiological age factor is computed as part of photosynthesis modeling. With PDT=23.3 physiological days, a=0.029 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=23.3 physiological days, a=0.029 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=23.3 physiological days, a=0.029 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: 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 a photosynthesis, the Physiological age factor is computed as part of photosynthesis modeling. With PDT=23.3 physiological days, a=0.029 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: 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 Physiological age factor is computed as part of photosynthesis modeling. With PDT=23.3 physiological days, a=0.029 dimensionless, determine 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=23.3 physiological days, a=0.029 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "gold", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF045"}
+{"answer": "1.232", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF002", "CMF014"], "effective_parameters": {"Biom_plant_yesterday": 436.8, "Coef_Rm": 0.0031, "DVI": 0.38, "Tmn": 11.6}, "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_t2_0000", "is_gold": true, "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_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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 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 with DVI correction", "mode": "branch", "parameters": {"Biom_plant_yesterday": 436.8, "Coef_Rm": 0.0031, "DVI": 0.38, "Tmn": 11.6}, "problem": "Calculate carbon-nitrogen cycling plant maintenance respiration in carbon-nitrogen cycling respiration with these inputs: Biom_plant_yesterday=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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: Calculate carbon-nitrogen cycling plant maintenance respiration in carbon-nitrogen cycling respiration with these inputs: Biom_plant_yesterday=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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: Calculate carbon-nitrogen cycling plant maintenance respiration in carbon-nitrogen cycling respiration with these inputs: Biom_plant_yesterday=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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: Calculate carbon-nitrogen cycling plant maintenance respiration in carbon-nitrogen cycling respiration with these inputs: Biom_plant_yesterday=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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: Calculate carbon-nitrogen cycling plant maintenance respiration in carbon-nitrogen cycling respiration with these inputs: Biom_plant_yesterday=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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: Calculate carbon-nitrogen cycling plant maintenance respiration in carbon-nitrogen cycling respiration with these inputs: Biom_plant_yesterday=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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 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 plant maintenance respiration in carbon-nitrogen cycling respiration with these inputs: Biom_plant_yesterday=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result 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=436.8 g C/m2, Coef_Rm=0.0031 g C/g C/s, Tmn=11.6 C, DVI=0.38 dimensionless.\nThe computed value is 1.232 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{1.232}.\n#### 1.232", "split": "gold", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "wrong_formula_id": "CMF002"}
+{"answer": "1.00", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.336, "old_value": 0.305, "parameter": "PS"}, "distractor_formula_ids": ["CMF029", "CMF026"], "effective_parameters": {"P": 9.75, "PS": 0.336, "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_sensitivity_t1_0001", "is_gold": true, "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: 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: 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: 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 quadratic", "mode": "sensitivity", "parameters": {"P": 9.75, "PS": 0.305, "Po": 10.5}, "problem": "Given current measurements P=9.75 h, Po=10.5 h, PS=0.305 dimensionless in phenology, what would the Relative photoperiod effectiveness be if PS were adjusted to 0.336? 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: 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 current measurements P=9.75 h, Po=10.5 h, PS=0.305 dimensionless in phenology, what would the Relative photoperiod effectiveness be if PS were adjusted to 0.336? 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: 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: CO2 response factor (二氧���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 current measurements P=9.75 h, Po=10.5 h, PS=0.305 dimensionless in phenology, what would the Relative photoperiod effectiveness be if PS were adjusted to 0.336? 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: 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 current measurements P=9.75 h, Po=10.5 h, PS=0.305 dimensionless in phenology, what would the Relative photoperiod effectiveness be if PS were adjusted to 0.336? 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: 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 current measurements P=9.75 h, Po=10.5 h, PS=0.305 dimensionless in phenology, what would the Relative photoperiod effectiveness be if PS were adjusted to 0.336? 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: 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 current measurements P=9.75 h, Po=10.5 h, PS=0.305 dimensionless in phenology, what would the Relative photoperiod effectiveness be if PS were adjusted to 0.336? 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 P=9.75 h, Po=10.5 h, PS=0.305 dimensionless in phenology, what would the Relative photoperiod effectiveness be if PS were adjusted to 0.336? Provide the 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=9.75 h, Po=10.5 h, PS=0.336 dimensionless.\nThe computed value is 1.00 fraction.\nTherefore, the final numeric answer is \\boxed{1.00}.\n#### 1.00", "split": "gold", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF029"}
+{"answer": "0.820", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF057", "CMF058"], "effective_parameters": {"Tbase": 5.4, "Tmean": 22.7, "Topt": 26.5}, "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_t0_0001", "is_gold": true, "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 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 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 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", "mode": "direct", "parameters": {"Tbase": 5.4, "Tmean": 22.7, "Topt": 26.5}, "problem": "A environmental-response formula system calculation estimates environmental-response temperature response. The measured inputs are Tmean=22.7 C, Tbase=5.4 C, Topt=26.5 C. What is the simple temperature 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_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 calculation estimates environmental-response temperature response. The measured inputs are Tmean=22.7 C, Tbase=5.4 C, Topt=26.5 C. What is the simple temperature 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 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 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 temperature response. The measured inputs are Tmean=22.7 C, Tbase=5.4 C, Topt=26.5 C. What is the simple temperature 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 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 calculation estimates environmental-response temperature response. The measured inputs are Tmean=22.7 C, Tbase=5.4 C, Topt=26.5 C. What is the simple temperature 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 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 calculation estimates environmental-response temperature response. The measured inputs are Tmean=22.7 C, Tbase=5.4 C, Topt=26.5 C. What is the simple temperature 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 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 calculation estimates environmental-response temperature response. The measured inputs are Tmean=22.7 C, Tbase=5.4 C, Topt=26.5 C. What is the simple temperature 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 temperature response. The measured inputs are Tmean=22.7 C, Tbase=5.4 C, Topt=26.5 C. What is the simple temperature response? Give 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=22.7 C, Tbase=5.4 C, Topt=26.5 C.\nThe computed value is 0.820 fraction.\nTherefore, the final numeric answer is \\boxed{0.820}.\n#### 0.820", "split": "gold", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF057"}
+{"answer": "13.47", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF026", "CMF039"], "effective_parameters": {"LAI": 4.31, "PAR": 14.13, "k": 0.71}, "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_t2_0000", "is_gold": true, "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: 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: 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: 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: 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": 1, "math_type": "Product with exponential saturation", "mode": "direct", "parameters": {"LAI": 4.31, "PAR": 14.13, "k": 0.71}, "problem": "In a photosynthesis, the Intercepted photosynthetically active radiation is computed as part of photosynthesis modeling. With PAR=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2, determine the 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: In a photosynthesis, the Intercepted photosynthetically active radiation is computed as part of photosynthesis modeling. With PAR=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2, determine the 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: 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: 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: 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 a photosynthesis, the Intercepted photosynthetically active radiation is computed as part of photosynthesis modeling. With PAR=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2, determine the 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: In a photosynthesis, the Intercepted photosynthetically active radiation is computed as part of photosynthesis modeling. With PAR=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2, determine the 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: In a photosynthesis, the Intercepted photosynthetically active radiation is computed as part of photosynthesis modeling. With PAR=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2, determine the 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: In a photosynthesis, the Intercepted photosynthetically active radiation is computed as part of photosynthesis modeling. With PAR=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2, determine the 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: 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 a photosynthesis, the Intercepted photosynthetically active radiation is computed as part of photosynthesis modeling. With PAR=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2, determine the 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=14.13 MJ/m2/day, k=0.71 dimensionless, LAI=4.31 m2/m2.\nThe computed value is 13.47 MJ/m2/day.\nTherefore, the final numeric answer is \\boxed{13.47}.\n#### 13.47", "split": "gold", "subject": "Photosynthesis", "template_id": 2, "wrong_formula_id": "CMF026"}
+{"answer": "0.28", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF049", "CMF052"], "effective_parameters": {"B_above": 15781.0, "PDT": 38.5, "Y_grain": 8168.0}, "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_t2_0000", "is_gold": true, "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: 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: 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: 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": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "parameters": {"B_above": 15781.0, "PDT": 38.5, "Y_grain": 8168.0}, "problem": "In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 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: 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: In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 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: 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: 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 dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 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: 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 dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 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: 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: In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 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: 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: In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 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: 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 dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 physiological days, determine 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=8168.0 kg/ha, B_above=15781.0 kg/ha, PDT=38.5 physiological days.\nThe computed value is 0.28 fraction.\nTherefore, the final numeric answer is \\boxed{0.28}.\n#### 0.28", "split": "gold", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF049"}
+{"answer": "13.06", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF048", "CMF042"], "effective_parameters": {"PS": 0.153, "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_t2_0001", "is_gold": true, "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: 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: 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: 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: 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": "Power response", "mode": "direct", "parameters": {"PS": 0.153, "Po": 10.5}, "problem": "In a phenology, the Critical daylength is computed as part of phenology modeling. With Po=10.5 h, PS=0.153 dimensionless, determine 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 a phenology, the Critical daylength is computed as part of phenology modeling. With Po=10.5 h, PS=0.153 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: 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: 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: 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 Critical daylength is computed as part of phenology modeling. With Po=10.5 h, PS=0.153 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 a phenology, the Critical daylength is computed as part of phenology modeling. With Po=10.5 h, PS=0.153 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 a phenology, the Critical daylength is computed as part of phenology modeling. With Po=10.5 h, PS=0.153 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 a phenology, the Critical daylength is computed as part of phenology modeling. With Po=10.5 h, PS=0.153 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: 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 Critical daylength is computed as part of phenology modeling. With Po=10.5 h, PS=0.153 dimensionless, determine 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=0.153 dimensionless.\nThe computed value is 13.06 h.\nTherefore, the final numeric answer is \\boxed{13.06}.\n#### 13.06", "split": "gold", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF048"}
+{"answer": "270.36", "benchmark_version": "cropmath-v1", "change": {"new_value": 180.9, "old_value": 201.0, "parameter": "SLAc"}, "distractor_formula_ids": ["CMF048", "CMF054"], "effective_parameters": {"GDD": 435.7, "SLAc": 180.9}, "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_t1_0000", "is_gold": true, "knowledge_distractor": "Candidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\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 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.", "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: 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": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "parameters": {"GDD": 435.7, "SLAc": 201.0}, "problem": "Given current measurements GDD=435.7 C day, SLAc=201.0 cm2/g in leaf area growth, what would the Specific leaf area be if SLAc were adjusted to 180.9? Provide 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: Given current measurements GDD=435.7 C day, SLAc=201.0 cm2/g in leaf area growth, what would the Specific leaf area be if SLAc were adjusted to 180.9? Provide 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: 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 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 current measurements GDD=435.7 C day, SLAc=201.0 cm2/g in leaf area growth, what would the Specific leaf area be if SLAc were adjusted to 180.9? Provide 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: Given current measurements GDD=435.7 C day, SLAc=201.0 cm2/g in leaf area growth, what would the Specific leaf area be if SLAc were adjusted to 180.9? Provide 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: Given current measurements GDD=435.7 C day, SLAc=201.0 cm2/g in leaf area growth, what would the Specific leaf area be if SLAc were adjusted to 180.9? Provide 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: Given current measurements GDD=435.7 C day, SLAc=201.0 cm2/g in leaf area growth, what would the Specific leaf area be if SLAc were adjusted to 180.9? Provide 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: 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 current measurements GDD=435.7 C day, SLAc=201.0 cm2/g in leaf area growth, what would the Specific leaf area be if SLAc were adjusted to 180.9? Provide 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=435.7 C day, SLAc=180.9 cm2/g.\nThe computed value is 270.36 cm2/g.\nTherefore, the final numeric answer is \\boxed{270.36}.\n#### 270.36", "split": "gold", "subject": "Leaf area growth", "template_id": 1, "wrong_formula_id": "CMF048"}
+{"answer": "0.006", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF017", "CMF021"], "effective_parameters": {"P": 5.374, "Wr": 1924.7, "t_soil": 26.3}, "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_0001", "is_gold": true, "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 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: 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": 5.374, "Wr": 1924.7, "t_soil": 26.3}, "problem": "Given P=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 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=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 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: 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 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: Given P=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 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=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 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=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 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=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 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=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 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=5.374 g CH4/m2/day, t_soil=26.3 C, Wr=1924.7 kg/ha.\nThe computed value is 0.006 g CH4/m2/day.\nTherefore, the final numeric answer is \\boxed{0.006}.\n#### 0.006", "split": "gold", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF017"}
+{"answer": "0.27", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.0054, "old_value": 0.0047, "parameter": "RP"}, "distractor_formula_ids": ["CMF044", "CMF041"], "effective_parameters": {"GDD": 150.5, "LAI0": 0.122, "RP": 0.0054}, "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_sensitivity_t0_0001", "is_gold": true, "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: 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: 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: 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: 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": 2, "math_type": "Exponential growth", "mode": "sensitivity", "parameters": {"GDD": 150.5, "LAI0": 0.122, "RP": 0.0047}, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.122 m2/m2, RP=0.0047 d-1, GDD=150.5 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0054. 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: 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=0.122 m2/m2, RP=0.0047 d-1, GDD=150.5 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0054. 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: 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: 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: 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 leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.122 m2/m2, RP=0.0047 d-1, GDD=150.5 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0054. 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=0.122 m2/m2, RP=0.0047 d-1, GDD=150.5 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0054. 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=0.122 m2/m2, RP=0.0047 d-1, GDD=150.5 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0054. 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=0.122 m2/m2, RP=0.0047 d-1, GDD=150.5 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0054. 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: 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 leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.122 m2/m2, RP=0.0047 d-1, GDD=150.5 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0054. 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=0.122 m2/m2, RP=0.0054 d-1, GDD=150.5 C day.\nThe computed value is 0.27 m2/m2.\nTherefore, the final numeric answer is \\boxed{0.27}.\n#### 0.27", "split": "gold", "subject": "Leaf area growth", "template_id": 0, "wrong_formula_id": "CMF044"}
+{"answer": "0.206", "benchmark_version": "cropmath-v1", "change": {"new_value": 3958.8, "old_value": 4398.7, "parameter": "W"}, "distractor_formula_ids": ["CMF022", "CMF025"], "effective_parameters": {"CH4RiceEfC": 0.233, "W": 3958.8, "Wmax": 10205.0}, "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_sensitivity_t2_0001", "is_gold": true, "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: 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 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 attenuation", "mode": "sensitivity", "parameters": {"CH4RiceEfC": 0.233, "W": 4398.7, "Wmax": 10205.0}, "problem": "In methane, we analyze how Plant methane transport factor responds to parameter changes. Baseline: CH4RiceEfC=0.233 fraction, W=4398.7 kg/ha, Wmax=10205.0 kg/ha. After changing W from 4398.7 to 3958.8, compute the new Plant methane transport 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: 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 methane, we analyze how Plant methane transport factor responds to parameter changes. Baseline: CH4RiceEfC=0.233 fraction, W=4398.7 kg/ha, Wmax=10205.0 kg/ha. After changing W from 4398.7 to 3958.8, compute the new Plant methane transport 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: 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: 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 methane, we analyze how Plant methane transport factor responds to parameter changes. Baseline: CH4RiceEfC=0.233 fraction, W=4398.7 kg/ha, Wmax=10205.0 kg/ha. After changing W from 4398.7 to 3958.8, compute the new Plant methane transport 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: 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 methane, we analyze how Plant methane transport factor responds to parameter changes. Baseline: CH4RiceEfC=0.233 fraction, W=4398.7 kg/ha, Wmax=10205.0 kg/ha. After changing W from 4398.7 to 3958.8, compute the new Plant methane transport 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: 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 methane, we analyze how Plant methane transport factor responds to parameter changes. Baseline: CH4RiceEfC=0.233 fraction, W=4398.7 kg/ha, Wmax=10205.0 kg/ha. After changing W from 4398.7 to 3958.8, compute the new Plant methane transport 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: 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 methane, we analyze how Plant methane transport factor responds to parameter changes. Baseline: CH4RiceEfC=0.233 fraction, W=4398.7 kg/ha, Wmax=10205.0 kg/ha. After changing W from 4398.7 to 3958.8, compute the new Plant methane transport 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: 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 Plant methane transport factor responds to parameter changes. Baseline: CH4RiceEfC=0.233 fraction, W=4398.7 kg/ha, Wmax=10205.0 kg/ha. After changing W from 4398.7 to 3958.8, compute the new Plant methane transport factor. Give 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=0.233 fraction, W=3958.8 kg/ha, Wmax=10205.0 kg/ha.\nThe computed value is 0.206 fraction.\nTherefore, the final numeric answer is \\boxed{0.206}.\n#### 0.206", "split": "gold", "subject": "Methane", "template_id": 2, "wrong_formula_id": "CMF022"}
+{"answer": "0.15", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF032", "CMF051"], "effective_parameters": {"PDT": 45.8}, "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_direct_t2_0001", "is_gold": true, "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: 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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\nUse the formula only when the problem context matches the stated physiological quantity.", "level": 2, "math_type": "Piecewise linear/exponential", "mode": "direct", "parameters": {"PDT": 45.8}, "problem": "In a dry matter partitioning, the Green leaf partitioning index is computed as part of dry matter partitioning modeling. With PDT=45.8 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: 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 a dry matter partitioning, the Green leaf partitioning index is computed as part of dry matter partitioning modeling. With PDT=45.8 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: 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: 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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a dry matter partitioning, the Green leaf partitioning index is computed as part of dry matter partitioning modeling. With PDT=45.8 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: 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 dry matter partitioning, the Green leaf partitioning index is computed as part of dry matter partitioning modeling. With PDT=45.8 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: 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 a dry matter partitioning, the Green leaf partitioning index is computed as part of dry matter partitioning modeling. With PDT=45.8 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: 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 a dry matter partitioning, the Green leaf partitioning index is computed as part of dry matter partitioning modeling. With PDT=45.8 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: 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<Tol: sin(pi/2*(T-Tb)/(Tol-Tb)); if Tol<=T<Tou: 1; if Tou<=T<Tmax: cos(pi/2*(T-Tou)/(Tmax-Tou)); 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- T: canopy temperature used by the response function; unit=C; typical range=[12.0, 42.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- Tol: lower optimum temperature; unit=C; typical range=[24.0, 24.0]\n- Tou: upper optimum temperature; unit=C; typical range=[34.0, 34.0]\n- Tmax: maximum temperature; unit=C; typical range=[45.0, 45.0]\nPhysiological category: Temperature response\nMathematical form: Piecewise trigonometric\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a dry matter partitioning, the Green leaf partitioning index is computed as part of dry matter partitioning modeling. With PDT=45.8 physiological days, determine 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=45.8 physiological days.\nThe computed value is 0.15 fraction.\nTherefore, the final numeric answer is \\boxed{0.15}.\n#### 0.15", "split": "gold", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF032"}
+{"answer": "43.148", "benchmark_version": "cropmath-v1", "change": {"new_value": 179.4, "old_value": 199.3, "parameter": "Coef_beta"}, "distractor_formula_ids": ["CMF012", "CMF005"], "effective_parameters": {"Coef_alfa": 12.68, "Coef_beta": 179.4, "Leaf_N": 4.01, "PAR": 1005.5}, "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_sensitivity_t0_0000", "is_gold": true, "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 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 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 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 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": "Michaelis-Menten response", "mode": "sensitivity", "parameters": {"Coef_alfa": 12.68, "Coef_beta": 199.3, "Leaf_N": 4.01, "PAR": 1005.5}, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Coef_alfa=12.68 umol CO2/gN/s, Coef_beta=199.3 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2. A field technician asks what the carbon-nitrogen cycling optimal photosynthetic rate would become if Coef_beta changed to 179.4. Report the new 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 involves sensitivity analysis. The current measurements are Coef_alfa=12.68 umol CO2/gN/s, Coef_beta=199.3 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2. A field technician asks what the carbon-nitrogen cycling optimal photosynthetic rate would become if Coef_beta changed to 179.4. Report the new 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 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 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 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 calculation involves sensitivity analysis. The current measurements are Coef_alfa=12.68 umol CO2/gN/s, Coef_beta=199.3 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2. A field technician asks what the carbon-nitrogen cycling optimal photosynthetic rate would become if Coef_beta changed to 179.4. Report the new 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 involves sensitivity analysis. The current measurements are Coef_alfa=12.68 umol CO2/gN/s, Coef_beta=199.3 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2. A field technician asks what the carbon-nitrogen cycling optimal photosynthetic rate would become if Coef_beta changed to 179.4. Report the new 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 involves sensitivity analysis. The current measurements are Coef_alfa=12.68 umol CO2/gN/s, Coef_beta=199.3 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2. A field technician asks what the carbon-nitrogen cycling optimal photosynthetic rate would become if Coef_beta changed to 179.4. Report the new 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 involves sensitivity analysis. The current measurements are Coef_alfa=12.68 umol CO2/gN/s, Coef_beta=199.3 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2. A field technician asks what the carbon-nitrogen cycling optimal photosynthetic rate would become if Coef_beta changed to 179.4. Report the new 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 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 calculation involves sensitivity analysis. The current measurements are Coef_alfa=12.68 umol CO2/gN/s, Coef_beta=199.3 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2. A field technician asks what the carbon-nitrogen cycling optimal photosynthetic rate would become if Coef_beta changed to 179.4. Report the new 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=12.68 umol CO2/gN/s, Coef_beta=179.4 umol/m2/s, PAR=1005.5 umol/m2/s, Leaf_N=4.01 gN/m2.\nThe computed value is 43.148 umol CO2/m2/s.\nTherefore, the final numeric answer is \\boxed{43.148}.\n#### 43.148", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "wrong_formula_id": "CMF012"}
+{"answer": "7.96", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.08, "old_value": 0.98, "parameter": "FCO2"}, "distractor_formula_ids": ["CMF030", "CMF047"], "effective_parameters": {"Am": 36.0, "FCO2": 1.08, "FPA": 0.49, "FT": 0.76, "NNI": 0.98, "WDF": 0.55}, "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_t1_0001", "is_gold": true, "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: 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���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: 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": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "parameters": {"Am": 36.0, "FCO2": 0.98, "FPA": 0.49, "FT": 0.76, "NNI": 0.98, "WDF": 0.55}, "problem": "Given current measurements Am=36.0 kg CO2/ha/h, FCO2=0.98 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction in photosynthesis, what would the Actual maximum assimilation rate be if FCO2 were adjusted to 1.08? Provide 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: Given current measurements Am=36.0 kg CO2/ha/h, FCO2=0.98 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction in photosynthesis, what would the Actual maximum assimilation rate be if FCO2 were adjusted to 1.08? Provide 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: 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: 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 Am=36.0 kg CO2/ha/h, FCO2=0.98 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction in photosynthesis, what would the Actual maximum assimilation rate be if FCO2 were adjusted to 1.08? Provide 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: Given current measurements Am=36.0 kg CO2/ha/h, FCO2=0.98 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction in photosynthesis, what would the Actual maximum assimilation rate be if FCO2 were adjusted to 1.08? Provide 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: Given current measurements Am=36.0 kg CO2/ha/h, FCO2=0.98 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction in photosynthesis, what would the Actual maximum assimilation rate be if FCO2 were adjusted to 1.08? Provide 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: Given current measurements Am=36.0 kg CO2/ha/h, FCO2=0.98 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction in photosynthesis, what would the Actual maximum assimilation rate be if FCO2 were adjusted to 1.08? Provide 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: 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 Am=36.0 kg CO2/ha/h, FCO2=0.98 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction in photosynthesis, what would the Actual maximum assimilation rate be if FCO2 were adjusted to 1.08? Provide 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=36.0 kg CO2/ha/h, FCO2=1.08 fraction, FPA=0.49 fraction, FT=0.76 fraction, NNI=0.98 fraction, WDF=0.55 fraction.\nThe computed value is 7.96 kg CO2/ha/h.\nTherefore, the final numeric answer is \\boxed{7.96}.\n#### 7.96", "split": "gold", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF030"}
+{"answer": "12.20", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.347, "old_value": 0.315, "parameter": "PS"}, "distractor_formula_ids": ["CMF031", "CMF036"], "effective_parameters": {"PS": 0.347, "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_0000", "is_gold": true, "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: 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: 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: 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: 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": "Power response", "mode": "sensitivity", "parameters": {"PS": 0.315, "Po": 10.5}, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.315 dimensionless. After changing PS from 0.315 to 0.347, 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=0.315 dimensionless. After changing PS from 0.315 to 0.347, 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: 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: 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: 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 phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.315 dimensionless. After changing PS from 0.315 to 0.347, 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=0.315 dimensionless. After changing PS from 0.315 to 0.347, 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=0.315 dimensionless. After changing PS from 0.315 to 0.347, 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=0.315 dimensionless. After changing PS from 0.315 to 0.347, 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: 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 phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.315 dimensionless. After changing PS from 0.315 to 0.347, 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=0.347 dimensionless.\nThe computed value is 12.20 h.\nTherefore, the final numeric answer is \\boxed{12.20}.\n#### 12.20", "split": "gold", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF031"}
+{"answer": "0.41", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF040", "CMF052"], "effective_parameters": {"B_above": 12870.0, "PDT": 42.5, "Y_grain": 6245.0}, "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_t2_0001", "is_gold": true, "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: 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: 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: 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: 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": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "parameters": {"B_above": 12870.0, "PDT": 42.5, "Y_grain": 6245.0}, "problem": "In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 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: 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: In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 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: 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: 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: 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 a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 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: 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 dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 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: 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: In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 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: 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: In a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 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: 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 a dry matter partitioning, the Panicle partitioning index is computed as part of dry matter partitioning modeling. With Y_grain=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 physiological days, determine 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=6245.0 kg/ha, B_above=12870.0 kg/ha, PDT=42.5 physiological days.\nThe computed value is 0.41 fraction.\nTherefore, the final numeric answer is \\boxed{0.41}.\n#### 0.41", "split": "gold", "subject": "Dry matter partitioning", "template_id": 2, "wrong_formula_id": "CMF040"}
+{"answer": "1.327", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF006", "CMF012"], "effective_parameters": {"CO2": 562.0, "Coef_B_CO2": 0.65}, "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_0000", "is_gold": true, "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 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 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 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 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": "Logarithmic", "mode": "direct", "parameters": {"CO2": 562.0, "Coef_B_CO2": 0.65}, "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=562.0 ppm, Coef_B_CO2=0.65 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=562.0 ppm, Coef_B_CO2=0.65 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 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 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 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 CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=562.0 ppm, Coef_B_CO2=0.65 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=562.0 ppm, Coef_B_CO2=0.65 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=562.0 ppm, Coef_B_CO2=0.65 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=562.0 ppm, Coef_B_CO2=0.65 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 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 CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=562.0 ppm, Coef_B_CO2=0.65 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=562.0 ppm, Coef_B_CO2=0.65 dimensionless.\nThe computed value is 1.327 fraction.\nTherefore, the final numeric answer is \\boxed{1.327}.\n#### 1.327", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF006"}
+{"answer": "0.01", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF040", "CMF051"], "effective_parameters": {"B_above": 15228.0, "PDT": 26.1, "Y_grain": 7014.0}, "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_t0_0000", "is_gold": true, "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: 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: 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": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "parameters": {"B_above": 15228.0, "PDT": 26.1, "Y_grain": 7014.0}, "problem": "A dry matter partitioning calculation estimates Panicle partitioning index. The measured inputs are Y_grain=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days. What is the panicle 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: 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: A dry matter partitioning calculation estimates Panicle partitioning index. The measured inputs are Y_grain=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days. What is the panicle 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: 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: 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 dry matter partitioning calculation estimates Panicle partitioning index. The measured inputs are Y_grain=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days. What is the panicle 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: 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 dry matter partitioning calculation estimates Panicle partitioning index. The measured inputs are Y_grain=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days. What is the panicle 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: 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: A dry matter partitioning calculation estimates Panicle partitioning index. The measured inputs are Y_grain=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days. What is the panicle 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: 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: A dry matter partitioning calculation estimates Panicle partitioning index. The measured inputs are Y_grain=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days. What is the panicle 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: 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 dry matter partitioning calculation estimates Panicle partitioning index. The measured inputs are Y_grain=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days. What is the panicle partitioning index? Give 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=7014.0 kg/ha, B_above=15228.0 kg/ha, PDT=26.1 physiological days.\nThe computed value is 0.01 fraction.\nTherefore, the final numeric answer is \\boxed{0.01}.\n#### 0.01", "split": "gold", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF040"}
+{"answer": "0.26", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF048", "CMF031"], "effective_parameters": {"PDT": 35.0}, "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_branch_t0_0001", "is_gold": true, "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: 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: 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: 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: 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": "Piecewise linear/exponential", "mode": "branch", "parameters": {"PDT": 35.0}, "problem": "A dry matter partitioning involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=35.0 physiological days. Determine the green leaf partitioning index 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: 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: A dry matter partitioning involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=35.0 physiological days. Determine the green leaf partitioning index 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: 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: 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: 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 dry matter partitioning involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=35.0 physiological days. Determine the green leaf partitioning index 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: 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 dry matter partitioning involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=35.0 physiological days. Determine the green leaf partitioning index 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: 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: A dry matter partitioning involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=35.0 physiological days. Determine the green leaf partitioning index 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: 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: A dry matter partitioning involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=35.0 physiological days. Determine the green leaf partitioning index 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 dry matter partitioning involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=35.0 physiological days. Determine the green leaf partitioning index and report the final 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=35.0 physiological days.\nThe computed value is 0.26 fraction.\nTherefore, the final numeric answer is \\boxed{0.26}.\n#### 0.26", "split": "gold", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF048"}
+{"answer": "1.21", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF041", "CMF028"], "effective_parameters": {"GDD": 298.3, "LAI0": 0.335, "RP": 0.0043}, "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_chain_2_t0_0000", "is_gold": true, "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: 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: 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: 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": "Exponential growth", "mode": "chain_2", "parameters": {"GDD": 298.3, "LAI0": 0.335, "RP": 0.0043}, "problem": "A leaf area growth module estimates Potential leaf area index using intermediate drivers. Use the provided inputs in sequence. The measurements are LAI0=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day. Compute the potential leaf area index and give the 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: 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 module estimates Potential leaf area index using intermediate drivers. Use the provided inputs in sequence. The measurements are LAI0=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day. Compute the potential leaf area index and give the 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: 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: 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 leaf area growth module estimates Potential leaf area index using intermediate drivers. Use the provided inputs in sequence. The measurements are LAI0=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day. Compute the potential leaf area index and give the 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 module estimates Potential leaf area index using intermediate drivers. Use the provided inputs in sequence. The measurements are LAI0=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day. Compute the potential leaf area index and give the 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 module estimates Potential leaf area index using intermediate drivers. Use the provided inputs in sequence. The measurements are LAI0=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day. Compute the potential leaf area index and give the 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 module estimates Potential leaf area index using intermediate drivers. Use the provided inputs in sequence. The measurements are LAI0=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day. Compute the potential leaf area index and give the 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: 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 leaf area growth module estimates Potential leaf area index using intermediate drivers. Use the provided inputs in sequence. The measurements are LAI0=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day. Compute the potential leaf area index and give the 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=0.335 m2/m2, RP=0.0043 d-1, GDD=298.3 C day.\nThe computed value is 1.21 m2/m2.\nTherefore, the final numeric answer is \\boxed{1.21}.\n#### 1.21", "split": "gold", "subject": "Leaf area growth", "template_id": 0, "wrong_formula_id": "CMF041"}
+{"answer": "410.21", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF049", "CMF044"], "effective_parameters": {"DTGA": 617.2, "RG": 121.4, "RM": 106.1, "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_chain_2_t0_0001", "is_gold": true, "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: 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: 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: 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: 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": "Subtraction and conversion", "mode": "chain_2", "parameters": {"DTGA": 617.2, "RG": 121.4, "RM": 106.1, "b": 0.05}, "problem": "A respiration module estimates Daily dry matter increment using intermediate drivers. Use the provided inputs in sequence. The measurements are DTGA=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction. Compute the daily dry matter increment and 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: 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: A respiration module estimates Daily dry matter increment using intermediate drivers. Use the provided inputs in sequence. The measurements are DTGA=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction. Compute the daily dry matter increment and 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: 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: 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: 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 respiration module estimates Daily dry matter increment using intermediate drivers. Use the provided inputs in sequence. The measurements are DTGA=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction. Compute the daily dry matter increment and 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: 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: A respiration module estimates Daily dry matter increment using intermediate drivers. Use the provided inputs in sequence. The measurements are DTGA=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction. Compute the daily dry matter increment and 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: 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: A respiration module estimates Daily dry matter increment using intermediate drivers. Use the provided inputs in sequence. The measurements are DTGA=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction. Compute the daily dry matter increment and 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: 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: A respiration module estimates Daily dry matter increment using intermediate drivers. Use the provided inputs in sequence. The measurements are DTGA=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction. Compute the daily dry matter increment and 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 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 module estimates Daily dry matter increment using intermediate drivers. Use the provided inputs in sequence. The measurements are DTGA=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction. Compute the daily dry matter increment and give the final 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=617.2 kg/ha/day, RM=106.1 kg/ha/day, RG=121.4 kg/ha/day, b=0.05 fraction.\nThe computed value is 410.21 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{410.21}.\n#### 410.21", "split": "gold", "subject": "Respiration", "template_id": 0, "wrong_formula_id": "CMF049"}
+{"answer": "1.000", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.372, "old_value": 0.338, "parameter": "Soil_Water"}, "distractor_formula_ids": ["CMF001", "CMF009"], "effective_parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.372, "Wilt_point": 0.08}, "family": "carbon_nitrogen_cycle", "formula": {"expression": "if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))", "id": "CMF014", "latex": "F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})", "name": "作物水分胁迫因子", "name_en": "carbon-nitrogen cycling crop water stress factor", "precision": 3, "unit": "fraction"}, "id": "CMF014_sensitivity_t1_0001", "is_gold": true, "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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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 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 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\nUse the formula only when the problem context matches the stated physiological quantity.", "knowledge_formula": "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_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]", "knowledge_formula_expr": "Formula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** 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 water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** 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 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": "Piecewise water response", "mode": "sensitivity", "parameters": {"Crop_Water_max": 0.42, "Crop_Water_min": 0.22, "Soil_Water": 0.338, "Wilt_point": 0.08}, "problem": "Given current measurements Soil_Water=0.338 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3 in carbon-nitrogen cycling water, what would the carbon-nitrogen cycling crop water stress factor be if Soil_Water were adjusted to 0.372? 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 = if Soil_Water<=Wilt_point: 0; if Soil_Water<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** 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 Soil_Water=0.338 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3 in carbon-nitrogen cycling water, what would the carbon-nitrogen cycling crop water stress factor be if Soil_Water were adjusted to 0.372? 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: 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water 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 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 Soil_Water=0.338 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3 in carbon-nitrogen cycling water, what would the carbon-nitrogen cycling crop water stress factor be if Soil_Water were adjusted to 0.372? 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: 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Soil_Water=0.338 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3 in carbon-nitrogen cycling water, what would the carbon-nitrogen cycling crop water stress factor be if Soil_Water were adjusted to 0.372? 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: 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\n\nQ: Given current measurements Soil_Water=0.338 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3 in carbon-nitrogen cycling water, what would the carbon-nitrogen cycling crop water stress factor be if Soil_Water were adjusted to 0.372? 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: 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** 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 Soil_Water=0.338 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3 in carbon-nitrogen cycling water, what would the carbon-nitrogen cycling crop water stress factor be if Soil_Water were adjusted to 0.372? 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: 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 Soil_Water=0.338 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3 in carbon-nitrogen cycling water, what would the carbon-nitrogen cycling crop water stress factor be if Soil_Water were adjusted to 0.372? Provide 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max)).\nSubstitute the relevant values:\nSoil_Water=0.372 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3.\nThe computed value is 1.000 fraction.\nTherefore, the final numeric answer is \\boxed{1.000}.\n#### 1.000", "split": "gold", "subject": "carbon-nitrogen cycling water", "template_id": 1, "wrong_formula_id": "CMF001"}
+{"answer": "475.26", "benchmark_version": "cropmath-v1", "change": {"new_value": 118.4, "old_value": 107.6, "parameter": "RM"}, "distractor_formula_ids": ["CMF045", "CMF036"], "effective_parameters": {"DTGA": 614.4, "RG": 44.5, "RM": 118.4, "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_t1_0001", "is_gold": true, "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: 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: 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 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: 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": "Subtraction and conversion", "mode": "sensitivity", "parameters": {"DTGA": 614.4, "RG": 44.5, "RM": 107.6, "b": 0.05}, "problem": "Given current measurements DTGA=614.4 kg/ha/day, RM=107.6 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction in respiration, what would the Daily dry matter increment be if RM were adjusted to 118.4? 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: 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 current measurements DTGA=614.4 kg/ha/day, RM=107.6 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction in respiration, what would the Daily dry matter increment be if RM were adjusted to 118.4? 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: 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: 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: 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: Given current measurements DTGA=614.4 kg/ha/day, RM=107.6 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction in respiration, what would the Daily dry matter increment be if RM were adjusted to 118.4? 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: 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 DTGA=614.4 kg/ha/day, RM=107.6 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction in respiration, what would the Daily dry matter increment be if RM were adjusted to 118.4? 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: 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 current measurements DTGA=614.4 kg/ha/day, RM=107.6 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction in respiration, what would the Daily dry matter increment be if RM were adjusted to 118.4? 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: 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 current measurements DTGA=614.4 kg/ha/day, RM=107.6 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction in respiration, what would the Daily dry matter increment be if RM were adjusted to 118.4? 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: 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 current measurements DTGA=614.4 kg/ha/day, RM=107.6 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction in respiration, what would the Daily dry matter increment be if RM were adjusted to 118.4? Provide 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=614.4 kg/ha/day, RM=118.4 kg/ha/day, RG=44.5 kg/ha/day, b=0.05 fraction.\nThe computed value is 475.26 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{475.26}.\n#### 475.26", "split": "gold", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF045"}
+{"answer": "0.69", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF040", "CMF029"], "effective_parameters": {"AWL": 3435.0, "SLA": 0.0002}, "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_t2_0001", "is_gold": true, "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: 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: 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: 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": 1, "math_type": "Product", "mode": "direct", "parameters": {"AWL": 3435.0, "SLA": 0.0002}, "problem": "In a leaf area growth, the Leaf area index from SLA is computed as part of leaf area growth modeling. With SLA=0.0002 ha/kg, AWL=3435.0 kg/ha, determine 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 a leaf area growth, the Leaf area index from SLA is computed as part of leaf area growth modeling. With SLA=0.0002 ha/kg, AWL=3435.0 kg/ha, determine 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: 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: 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 a leaf area growth, the Leaf area index from SLA is computed as part of leaf area growth modeling. With SLA=0.0002 ha/kg, AWL=3435.0 kg/ha, determine 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 a leaf area growth, the Leaf area index from SLA is computed as part of leaf area growth modeling. With SLA=0.0002 ha/kg, AWL=3435.0 kg/ha, determine 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 a leaf area growth, the Leaf area index from SLA is computed as part of leaf area growth modeling. With SLA=0.0002 ha/kg, AWL=3435.0 kg/ha, determine 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 a leaf area growth, the Leaf area index from SLA is computed as part of leaf area growth modeling. With SLA=0.0002 ha/kg, AWL=3435.0 kg/ha, determine 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: 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 a leaf area growth, the Leaf area index from SLA is computed as part of leaf area growth modeling. With SLA=0.0002 ha/kg, AWL=3435.0 kg/ha, determine 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=0.0002 ha/kg, AWL=3435.0 kg/ha.\nThe computed value is 0.69 m2/m2.\nTherefore, the final numeric answer is \\boxed{0.69}.\n#### 0.69", "split": "gold", "subject": "Leaf area growth", "template_id": 2, "wrong_formula_id": "CMF040"}
+{"answer": "1.413", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF062", "CMF055"], "effective_parameters": {"CO2": 608.0, "CO2_RUE": 0.16}, "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_t2_0000", "is_gold": true, "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 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 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": 4, "math_type": "Piecewise linear cap", "mode": "branch", "parameters": {"CO2": 608.0, "CO2_RUE": 0.16}, "problem": "Calculate environmental-response CO2 response in environmental-response formula system with these inputs: CO2=608.0 ppm, CO2_RUE=0.16 fraction. 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_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: Calculate environmental-response CO2 response in environmental-response formula system with these inputs: CO2=608.0 ppm, CO2_RUE=0.16 fraction. 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 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 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: Calculate environmental-response CO2 response in environmental-response formula system with these inputs: CO2=608.0 ppm, CO2_RUE=0.16 fraction. 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 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: Calculate environmental-response CO2 response in environmental-response formula system with these inputs: CO2=608.0 ppm, CO2_RUE=0.16 fraction. 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 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: Calculate environmental-response CO2 response in environmental-response formula system with these inputs: CO2=608.0 ppm, CO2_RUE=0.16 fraction. 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 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: Calculate environmental-response CO2 response in environmental-response formula system with these inputs: CO2=608.0 ppm, CO2_RUE=0.16 fraction. 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 CO2 response in environmental-response formula system with these inputs: CO2=608.0 ppm, CO2_RUE=0.16 fraction. 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 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=608.0 ppm, CO2_RUE=0.16 fraction.\nThe computed value is 1.413 fraction.\nTherefore, the final numeric answer is \\boxed{1.413}.\n#### 1.413", "split": "gold", "subject": "environmental-response formula system", "template_id": 2, "wrong_formula_id": "CMF062"}
+{"answer": "1.376", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF017", "CMF025"], "effective_parameters": {"OMS": 863.8, "SI": 1.02, "TI": 1.052, "WI": 0.495}, "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_t1_0001", "is_gold": true, "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: 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���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: 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": "Multiplicative", "mode": "chain_2", "parameters": {"OMS": 863.8, "SI": 1.02, "TI": 1.052, "WI": 0.495}, "problem": "Given WI=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2, compute Soil organic matter carbon decomposition in methane. 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: 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: Given WI=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2, compute Soil organic matter carbon decomposition in methane. 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: 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: 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: Given WI=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2, compute Soil organic matter carbon decomposition in methane. 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: 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 WI=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2, compute Soil organic matter carbon decomposition in methane. 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: 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: Given WI=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2, compute Soil organic matter carbon decomposition in methane. 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: 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: Given WI=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2, compute Soil organic matter carbon decomposition in methane. 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: 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 WI=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2, compute Soil organic matter carbon decomposition in methane. 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 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=0.495 fraction, SI=1.02 fraction, TI=1.052 fraction, OMS=863.8 g C/m2.\nThe computed value is 1.376 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{1.376}.\n#### 1.376", "split": "gold", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF017"}
+{"answer": "0.70", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF043", "CMF048"], "effective_parameters": {"PDT": 7.3}, "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_direct_t0_0000", "is_gold": true, "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: 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: 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: 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": 1, "math_type": "Quadratic polynomial", "mode": "direct", "parameters": {"PDT": 7.3}, "problem": "A dry matter partitioning calculation estimates Shoot partitioning index. The measured inputs are PDT=7.3 physiological days. What is the shoot 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: 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 estimates Shoot partitioning index. The measured inputs are PDT=7.3 physiological days. What is the shoot 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: 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: 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 calculation estimates Shoot partitioning index. The measured inputs are PDT=7.3 physiological days. What is the shoot 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: 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 estimates Shoot partitioning index. The measured inputs are PDT=7.3 physiological days. What is the shoot 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: 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 estimates Shoot partitioning index. The measured inputs are PDT=7.3 physiological days. What is the shoot 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: 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 estimates Shoot partitioning index. The measured inputs are PDT=7.3 physiological days. What is the shoot 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: 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 estimates Shoot partitioning index. The measured inputs are PDT=7.3 physiological days. What is the shoot partitioning index? Give the 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=7.3 physiological days.\nThe computed value is 0.70 fraction.\nTherefore, the final numeric answer is \\boxed{0.70}.\n#### 0.70", "split": "gold", "subject": "Dry matter partitioning", "template_id": 0, "wrong_formula_id": "CMF043"}
+{"answer": "28.04", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF022", "CMF016"], "effective_parameters": {"Tair": 31.1}, "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_0001", "is_gold": true, "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: 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: 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": "Linear", "mode": "direct", "parameters": {"Tair": 31.1}, "problem": "Given Tair=31.1 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=31.1 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: 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: 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=31.1 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=31.1 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=31.1 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=31.1 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: 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 Tair=31.1 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=31.1 C.\nThe computed value is 28.04 C.\nTherefore, the final numeric answer is \\boxed{28.04}.\n#### 28.04", "split": "gold", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF022"}
+{"answer": "295.16", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF037", "CMF029"], "effective_parameters": {"DTGA": 460.2, "RG": 93.0, "RM": 86.8, "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_t1_0000", "is_gold": true, "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 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: 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 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: 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": "Subtraction and conversion", "mode": "chain_2", "parameters": {"DTGA": 460.2, "RG": 93.0, "RM": 86.8, "b": 0.05}, "problem": "Given DTGA=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction, compute Daily dry matter increment 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: 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=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction, compute Daily dry matter increment 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: 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 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: 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 DTGA=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction, compute Daily dry matter increment 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: 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=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction, compute Daily dry matter increment 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: 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=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction, compute Daily dry matter increment 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: 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=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction, compute Daily dry matter increment 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: 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 DTGA=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction, compute Daily dry matter increment 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 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=460.2 kg/ha/day, RM=86.8 kg/ha/day, RG=93.0 kg/ha/day, b=0.05 fraction.\nThe computed value is 295.16 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{295.16}.\n#### 295.16", "split": "gold", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF037"}
+{"answer": "0.18", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF044", "CMF034"], "effective_parameters": {"P": 12.42, "PS": 0.223, "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_chain_2_t2_0000", "is_gold": true, "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: 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: 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: 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: 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": 4, "math_type": "Piecewise quadratic", "mode": "chain_2", "parameters": {"P": 12.42, "PS": 0.223, "Po": 10.5}, "problem": "In phenology, the Relative photoperiod effectiveness is obtained by chaining intermediate calculations. With inputs P=12.42 h, Po=10.5 h, PS=0.223 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: 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: In phenology, the Relative photoperiod effectiveness is obtained by chaining intermediate calculations. With inputs P=12.42 h, Po=10.5 h, PS=0.223 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: 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: 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: 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 phenology, the Relative photoperiod effectiveness is obtained by chaining intermediate calculations. With inputs P=12.42 h, Po=10.5 h, PS=0.223 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: 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 phenology, the Relative photoperiod effectiveness is obtained by chaining intermediate calculations. With inputs P=12.42 h, Po=10.5 h, PS=0.223 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: 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: In phenology, the Relative photoperiod effectiveness is obtained by chaining intermediate calculations. With inputs P=12.42 h, Po=10.5 h, PS=0.223 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: 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: In phenology, the Relative photoperiod effectiveness is obtained by chaining intermediate calculations. With inputs P=12.42 h, Po=10.5 h, PS=0.223 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: 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 phenology, the Relative photoperiod effectiveness is obtained by chaining intermediate calculations. With inputs P=12.42 h, Po=10.5 h, PS=0.223 dimensionless, determine the 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=12.42 h, Po=10.5 h, PS=0.223 dimensionless.\nThe computed value is 0.18 fraction.\nTherefore, the final numeric answer is \\boxed{0.18}.\n#### 0.18", "split": "gold", "subject": "Phenology", "template_id": 2, "wrong_formula_id": "CMF044"}
+{"answer": "0.624", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF056", "CMF062"], "effective_parameters": {"Tbase": 6.9, "Tmean": 16.7, "Topt": 22.6}, "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_t1_0001", "is_gold": true, "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 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 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 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": 4, "math_type": "Piecewise linear", "mode": "branch", "parameters": {"Tbase": 6.9, "Tmean": 16.7, "Topt": 22.6}, "problem": "For environmental-response formula system, compute the environmental-response temperature response. The inputs are Tmean=16.7 C, Tbase=6.9 C, Topt=22.6 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: 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: For environmental-response formula system, compute the environmental-response temperature response. The inputs are Tmean=16.7 C, Tbase=6.9 C, Topt=22.6 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: 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 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 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: For environmental-response formula system, compute the environmental-response temperature response. The inputs are Tmean=16.7 C, Tbase=6.9 C, Topt=22.6 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: 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: For environmental-response formula system, compute the environmental-response temperature response. The inputs are Tmean=16.7 C, Tbase=6.9 C, Topt=22.6 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: 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: For environmental-response formula system, compute the environmental-response temperature response. The inputs are Tmean=16.7 C, Tbase=6.9 C, Topt=22.6 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: 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: For environmental-response formula system, compute the environmental-response temperature response. The inputs are Tmean=16.7 C, Tbase=6.9 C, Topt=22.6 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: 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 temperature response. The inputs are Tmean=16.7 C, Tbase=6.9 C, Topt=22.6 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 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=16.7 C, Tbase=6.9 C, Topt=22.6 C.\nThe computed value is 0.624 fraction.\nTherefore, the final numeric answer is \\boxed{0.624}.\n#### 0.624", "split": "gold", "subject": "environmental-response formula system", "template_id": 1, "wrong_formula_id": "CMF056"}
+{"answer": "1.100", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF005", "CMF014"], "effective_parameters": {"Coef_K_Decom": 0.0035, "Coef_K_Trans": 0.349, "F_notill": 1.09, "F_pH": 0.932, "F_temp": 2.474, "F_texture": 0.933, "F_water": 0.676, "Pool": 304.5}, "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_t2_0000", "is_gold": true, "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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\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 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 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": "Chained multiplicative decomposition", "mode": "direct", "parameters": {"Coef_K_Decom": 0.0035, "Coef_K_Trans": 0.349, "F_notill": 1.09, "F_pH": 0.932, "F_temp": 2.474, "F_texture": 0.933, "F_water": 0.676, "Pool": 304.5}, "problem": "In a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is computed as part of carbon-nitrogen cycling soil carbon modeling. With Pool=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is computed as part of carbon-nitrogen cycling soil carbon modeling. With Pool=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 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 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<Crop_Water_min: (Soil_Water-Wilt_point)/(Crop_Water_min-Wilt_point); if Soil_Water<Crop_Water_max: 1; else: 0.5*(1+(1-Soil_Water)/(1-Crop_Water_max))\nOperator notes: * means multiplication, / means division, ** or ^ means 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, 0.58]\n- Crop_Water_max: upper optimal crop water content; unit=cm3/cm3; typical range=[0.42, 0.42]\n- Crop_Water_min: lower optimal crop water content; unit=cm3/cm3; typical range=[0.22, 0.22]\n- Wilt_point: wilting point; unit=cm3/cm3; typical range=[0.08, 0.08]\nPhysiological category: carbon-nitrogen cycling water\nMathematical form: Piecewise water response\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 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 soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is computed as part of carbon-nitrogen cycling soil carbon modeling. With Pool=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is computed as part of carbon-nitrogen cycling soil carbon modeling. With Pool=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is computed as part of carbon-nitrogen cycling soil carbon modeling. With Pool=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 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 a carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is computed as part of carbon-nitrogen cycling soil carbon modeling. With Pool=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 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 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 soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is computed as part of carbon-nitrogen cycling soil carbon modeling. With Pool=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 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=304.5 g C/m2, Coef_K_Decom=0.0035 d-1, F_temp=2.474 fraction, F_water=0.676 fraction, F_texture=0.933 fraction, F_pH=0.932 fraction, F_notill=1.09 fraction, Coef_K_Trans=0.349 fraction.\nThe computed value is 1.100 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{1.100}.\n#### 1.100", "split": "gold", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "wrong_formula_id": "CMF005"}
+{"answer": "1.37", "benchmark_version": "cropmath-v1", "change": {"new_value": 0.81, "old_value": 0.74, "parameter": "beta"}, "distractor_formula_ids": ["CMF052", "CMF037"], "effective_parameters": {"Cx": 540.0, "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_t1_0001", "is_gold": true, "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: 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: 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: 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": 2, "math_type": "Logarithmic", "mode": "sensitivity", "parameters": {"Cx": 540.0, "beta": 0.74}, "problem": "Given current measurements Cx=540.0 ppm, beta=0.74 dimensionless in photosynthesis, what would the CO2 response factor be if beta were adjusted to 0.81? 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: 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 current measurements Cx=540.0 ppm, beta=0.74 dimensionless in photosynthesis, what would the CO2 response factor be if beta were adjusted to 0.81? 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: 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: 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 Cx=540.0 ppm, beta=0.74 dimensionless in photosynthesis, what would the CO2 response factor be if beta were adjusted to 0.81? 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: 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 Cx=540.0 ppm, beta=0.74 dimensionless in photosynthesis, what would the CO2 response factor be if beta were adjusted to 0.81? 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: 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 current measurements Cx=540.0 ppm, beta=0.74 dimensionless in photosynthesis, what would the CO2 response factor be if beta were adjusted to 0.81? 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: 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 current measurements Cx=540.0 ppm, beta=0.74 dimensionless in photosynthesis, what would the CO2 response factor be if beta were adjusted to 0.81? 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: 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 current measurements Cx=540.0 ppm, beta=0.74 dimensionless in photosynthesis, what would the CO2 response factor be if beta were adjusted to 0.81? Provide 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=540.0 ppm, beta=0.81 dimensionless.\nThe computed value is 1.37 fraction.\nTherefore, the final numeric answer is \\boxed{1.37}.\n#### 1.37", "split": "gold", "subject": "Photosynthesis", "template_id": 1, "wrong_formula_id": "CMF052"}
+{"answer": "0.86", "benchmark_version": "cropmath-v1", "change": {"new_value": 31.4, "old_value": 34.9, "parameter": "PDT"}, "distractor_formula_ids": ["CMF041", "CMF042"], "effective_parameters": {"PDT": 31.4}, "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_t1_0001", "is_gold": true, "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: 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: 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: 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: 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": "Quadratic polynomial", "mode": "sensitivity", "parameters": {"PDT": 34.9}, "problem": "Given current measurements PDT=34.9 physiological days in dry matter partitioning, what would the Shoot partitioning index be if PDT were adjusted to 31.4? 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: 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: Given current measurements PDT=34.9 physiological days in dry matter partitioning, what would the Shoot partitioning index be if PDT were adjusted to 31.4? 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: 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: 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: 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 current measurements PDT=34.9 physiological days in dry matter partitioning, what would the Shoot partitioning index be if PDT were adjusted to 31.4? 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: 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=34.9 physiological days in dry matter partitioning, what would the Shoot partitioning index be if PDT were adjusted to 31.4? 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: 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: Given current measurements PDT=34.9 physiological days in dry matter partitioning, what would the Shoot partitioning index be if PDT were adjusted to 31.4? 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: 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: Given current measurements PDT=34.9 physiological days in dry matter partitioning, what would the Shoot partitioning index be if PDT were adjusted to 31.4? 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: 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 current measurements PDT=34.9 physiological days in dry matter partitioning, what would the Shoot partitioning index be if PDT were adjusted to 31.4? Provide 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=31.4 physiological days.\nThe computed value is 0.86 fraction.\nTherefore, the final numeric answer is \\boxed{0.86}.\n#### 0.86", "split": "gold", "subject": "Dry matter partitioning", "template_id": 1, "wrong_formula_id": "CMF041"}
+{"answer": "2.995", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF006", "CMF012"], "effective_parameters": {"Biom_plant_yesterday": 215.0, "Coef_Rm": 0.0038, "DVI": 0.54, "Tmn": 31.7}, "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_0001", "is_gold": true, "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 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 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 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 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": "Q10 with DVI correction", "mode": "direct", "parameters": {"Biom_plant_yesterday": 215.0, "Coef_Rm": 0.0038, "DVI": 0.54, "Tmn": 31.7}, "problem": "A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? 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: 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=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 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 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 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 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 respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 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=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 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=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 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=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 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 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 respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 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:"}, "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=215.0 g C/m2, Coef_Rm=0.0038 g C/g C/s, Tmn=31.7 C, DVI=0.54 dimensionless.\nThe computed value is 2.995 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{2.995}.\n#### 2.995", "split": "gold", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "wrong_formula_id": "CMF006"}
+{"answer": "28.928", "benchmark_version": "cropmath-v1", "change": {"new_value": 15.72, "old_value": 13.67, "parameter": "Coef_alfa"}, "distractor_formula_ids": ["CMF011", "CMF001"], "effective_parameters": {"Coef_alfa": 15.72, "Coef_beta": 206.5, "Leaf_N": 2.39, "PAR": 691.2}, "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_sensitivity_t2_0001", "is_gold": true, "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 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 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 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": "Michaelis-Menten response", "mode": "sensitivity", "parameters": {"Coef_alfa": 13.67, "Coef_beta": 206.5, "Leaf_N": 2.39, "PAR": 691.2}, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling optimal photosynthetic rate responds to parameter changes. Baseline: Coef_alfa=13.67 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2. After changing Coef_alfa from 13.67 to 15.72, compute the new carbon-nitrogen cycling optimal photosynthetic rate. Give 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, we analyze how carbon-nitrogen cycling optimal photosynthetic rate responds to parameter changes. Baseline: Coef_alfa=13.67 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2. After changing Coef_alfa from 13.67 to 15.72, compute the new carbon-nitrogen cycling optimal photosynthetic rate. Give 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 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 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 optimal photosynthetic rate responds to parameter changes. Baseline: Coef_alfa=13.67 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2. After changing Coef_alfa from 13.67 to 15.72, compute the new carbon-nitrogen cycling optimal photosynthetic rate. Give 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, we analyze how carbon-nitrogen cycling optimal photosynthetic rate responds to parameter changes. Baseline: Coef_alfa=13.67 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2. After changing Coef_alfa from 13.67 to 15.72, compute the new carbon-nitrogen cycling optimal photosynthetic rate. Give 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, we analyze how carbon-nitrogen cycling optimal photosynthetic rate responds to parameter changes. Baseline: Coef_alfa=13.67 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2. After changing Coef_alfa from 13.67 to 15.72, compute the new carbon-nitrogen cycling optimal photosynthetic rate. Give 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, we analyze how carbon-nitrogen cycling optimal photosynthetic rate responds to parameter changes. Baseline: Coef_alfa=13.67 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2. After changing Coef_alfa from 13.67 to 15.72, compute the new carbon-nitrogen cycling optimal photosynthetic rate. Give 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 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 carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling optimal photosynthetic rate responds to parameter changes. Baseline: Coef_alfa=13.67 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2. After changing Coef_alfa from 13.67 to 15.72, compute the new carbon-nitrogen cycling optimal photosynthetic rate. Give 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=15.72 umol CO2/gN/s, Coef_beta=206.5 umol/m2/s, PAR=691.2 umol/m2/s, Leaf_N=2.39 gN/m2.\nThe computed value is 28.928 umol CO2/m2/s.\nTherefore, the final numeric answer is \\boxed{28.928}.\n#### 28.928", "split": "gold", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "wrong_formula_id": "CMF011"}
+{"answer": "1.354", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF015", "CMF018"], "effective_parameters": {"OMN": 129.9, "SI": 1.2, "TI": 1.2, "WI": 0.268}, "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_direct_t1_0001", "is_gold": true, "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: 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: 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: 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": "Multiplicative", "mode": "direct", "parameters": {"OMN": 129.9, "SI": 1.2, "TI": 1.2, "WI": 0.268}, "problem": "Given WI=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2, calculate the Fresh organic matter carbon decomposition for a methane. 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: 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 WI=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2, calculate the Fresh organic matter carbon decomposition for a methane. 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: 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: 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 WI=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2, calculate the Fresh organic matter carbon decomposition for a methane. 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: 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 WI=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2, calculate the Fresh organic matter carbon decomposition for a methane. 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: 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 WI=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2, calculate the Fresh organic matter carbon decomposition for a methane. 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: 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 WI=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2, calculate the Fresh organic matter carbon decomposition for a methane. 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: 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 WI=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2, calculate the Fresh organic matter carbon decomposition for a methane. The result should be 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=0.268 fraction, SI=1.2 fraction, TI=1.2 fraction, OMN=129.9 g C/m2.\nThe computed value is 1.354 g C/m2/day.\nTherefore, the final numeric answer is \\boxed{1.354}.\n#### 1.354", "split": "gold", "subject": "Methane", "template_id": 1, "wrong_formula_id": "CMF015"}
+{"answer": "0.358", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF035", "CMF041"], "effective_parameters": {"T": 39.8, "TS": 1.66, "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_0001", "is_gold": true, "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: 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: 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: 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: 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": "Beta temperature response", "mode": "direct", "parameters": {"T": 39.8, "TS": 1.66, "Tb": 10.0, "Tm": 42.0, "To": 30.0}, "problem": "Given T=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 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=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 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: 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: 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: 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 T=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 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=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 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=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 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=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 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: 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 T=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 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=39.8 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.66 dimensionless.\nThe computed value is 0.358 fraction.\nTherefore, the final numeric answer is \\boxed{0.358}.\n#### 0.358", "split": "gold", "subject": "Phenology", "template_id": 1, "wrong_formula_id": "CMF035"}
+{"answer": "14.91", "benchmark_version": "cropmath-v1", "change": null, "distractor_formula_ids": ["CMF037", "CMF035"], "effective_parameters": {"ABIOMASS": 6712.0, "PDT": 41.8, "Q10": 1.81, "T": 16.0, "To": 29.4}, "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_t1_0000", "is_gold": true, "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: 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: 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: 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": 6712.0, "PDT": 41.8, "Q10": 1.81, "T": 16.0, "To": 29.4}, "problem": "Given PDT=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 C, calculate the Maintenance respiration 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: 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=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 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: 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: 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: 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=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 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=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 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=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 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=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 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: 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=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 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=41.8 physiological days, ABIOMASS=6712.0 kg/ha, Q10=1.81 dimensionless, T=16.0 C, To=29.4 C.\nThe computed value is 14.91 kg/ha/day.\nTherefore, the final numeric answer is \\boxed{14.91}.\n#### 14.91", "split": "gold", "subject": "Respiration", "template_id": 1, "wrong_formula_id": "CMF037"}
+{"answer": "0.000", "benchmark_version": "cropmath-v1", "change": {"new_value": 5.9, "old_value": 5.4, "parameter": "Tmean"}, "distractor_formula_ids": ["CMF056", "CMF057"], "effective_parameters": {"Tbase": 7.5, "Tmean": 5.9, "Topt": 24.8}, "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_sensitivity_t0_0001", "is_gold": true, "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 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 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": 7.5, "Tmean": 5.4, "Topt": 24.8}, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are Tmean=5.4 C, Tbase=7.5 C, Topt=24.8 C. A field technician asks what the simple temperature response would become if Tmean changed to 5.9. 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: 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 calculation involves sensitivity analysis. The current measurements are Tmean=5.4 C, Tbase=7.5 C, Topt=24.8 C. A field technician asks what the simple temperature response would become if Tmean changed to 5.9. 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: 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 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: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are Tmean=5.4 C, Tbase=7.5 C, Topt=24.8 C. A field technician asks what the simple temperature response would become if Tmean changed to 5.9. 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: 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 calculation involves sensitivity analysis. The current measurements are Tmean=5.4 C, Tbase=7.5 C, Topt=24.8 C. A field technician asks what the simple temperature response would become if Tmean changed to 5.9. 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: 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 calculation involves sensitivity analysis. The current measurements are Tmean=5.4 C, Tbase=7.5 C, Topt=24.8 C. A field technician asks what the simple temperature response would become if Tmean changed to 5.9. 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: 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 calculation involves sensitivity analysis. The current measurements are Tmean=5.4 C, Tbase=7.5 C, Topt=24.8 C. A field technician asks what the simple temperature response would become if Tmean changed to 5.9. 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: 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 involves sensitivity analysis. The current measurements are Tmean=5.4 C, Tbase=7.5 C, Topt=24.8 C. A field technician asks what the simple temperature response would become if Tmean changed to 5.9. Report the new 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=5.9 C, Tbase=7.5 C, Topt=24.8 C.\nThe computed value is 0.000 fraction.\nTherefore, the final numeric answer is \\boxed{0.000}.\n#### 0.000", "split": "gold", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF056"}
+{"answer": "0.425", "benchmark_version": "cropmath-v1", "change": {"new_value": 1.69, "old_value": 1.88, "parameter": "ARID"}, "distractor_formula_ids": ["CMF058", "CMF059"], "effective_parameters": {"ARID": 1.69, "S_Water": 0.34}, "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_0000", "is_gold": true, "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 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 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": 2, "math_type": "Maximum linear", "mode": "sensitivity", "parameters": {"ARID": 1.88, "S_Water": 0.34}, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.88 dimensionless, S_Water=0.34 dimensionless. A field technician asks what the simple water stress response would become if ARID changed to 1.69. 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=1.88 dimensionless, S_Water=0.34 dimensionless. A field technician asks what the simple water stress response would become if ARID changed to 1.69. 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 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 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=1.88 dimensionless, S_Water=0.34 dimensionless. A field technician asks what the simple water stress response would become if ARID changed to 1.69. 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=1.88 dimensionless, S_Water=0.34 dimensionless. A field technician asks what the simple water stress response would become if ARID changed to 1.69. 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=1.88 dimensionless, S_Water=0.34 dimensionless. A field technician asks what the simple water stress response would become if ARID changed to 1.69. 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=1.88 dimensionless, S_Water=0.34 dimensionless. A field technician asks what the simple water stress response would become if ARID changed to 1.69. 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 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 involves sensitivity analysis. The current measurements are ARID=1.88 dimensionless, S_Water=0.34 dimensionless. A field technician asks what the simple water stress response would become if ARID changed to 1.69. 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=1.69 dimensionless, S_Water=0.34 dimensionless.\nThe computed value is 0.425 fraction.\nTherefore, the final numeric answer is \\boxed{0.425}.\n#### 0.425", "split": "gold", "subject": "environmental-response formula system", "template_id": 0, "wrong_formula_id": "CMF058"}