diff --git "a/data/eval_prompts/dev.jsonl" "b/data/eval_prompts/dev.jsonl" new file mode 100644--- /dev/null +++ "b/data/eval_prompts/dev.jsonl" @@ -0,0 +1,1512 @@ +{"answer": "6.247", "answer_float": 6.247, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "6.247", "answer_float": 6.247, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "6.247", "answer_float": 6.247, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "6.247", "answer_float": 6.247, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "6.247", "answer_float": 6.247, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "6.247", "answer_float": 6.247, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling respiration calculation estimates carbon-nitrogen cycling plant maintenance respiration. The measured inputs are Biom_plant_yesterday=704.9 g C/m2, Coef_Rm=0.0039 g C/g C/s, Tmn=24.8 C, DVI=0.48 dimensionless. What is the carbon-nitrogen cycling plant maintenance respiration? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "72.07", "answer_float": 72.07, "change": {"new_value": 30.0, "old_value": 29.3, "parameter": "To"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_sensitivity_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Product with Q10 temperature response", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "72.07", "answer_float": 72.07, "change": {"new_value": 30.0, "old_value": 29.3, "parameter": "To"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_sensitivity_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Product with Q10 temperature response", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "72.07", "answer_float": 72.07, "change": {"new_value": 30.0, "old_value": 29.3, "parameter": "To"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Product with Q10 temperature response", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "72.07", "answer_float": 72.07, "change": {"new_value": 30.0, "old_value": 29.3, "parameter": "To"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Product with Q10 temperature response", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "72.07", "answer_float": 72.07, "change": {"new_value": 30.0, "old_value": 29.3, "parameter": "To"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Product with Q10 temperature response", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "72.07", "answer_float": 72.07, "change": {"new_value": 30.0, "old_value": 29.3, "parameter": "To"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Product with Q10 temperature response", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Maintenance respiration responds to parameter changes. Baseline: PDT=7.0 physiological days, ABIOMASS=13378.0 kg/ha, Q10=2.18 dimensionless, T=24.3 C, To=29.3 C. After changing To from 29.3 to 30.0, compute the new Maintenance respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "28.34", "answer_float": 28.34, "change": {"new_value": 31.5, "old_value": 28.6, "parameter": "Tair"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "28.34", "answer_float": 28.34, "change": {"new_value": 31.5, "old_value": 28.6, "parameter": "Tair"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "28.34", "answer_float": 28.34, "change": {"new_value": 31.5, "old_value": 28.6, "parameter": "Tair"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "28.34", "answer_float": 28.34, "change": {"new_value": 31.5, "old_value": 28.6, "parameter": "Tair"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "28.34", "answer_float": 28.34, "change": {"new_value": 31.5, "old_value": 28.6, "parameter": "Tair"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "28.34", "answer_float": 28.34, "change": {"new_value": 31.5, "old_value": 28.6, "parameter": "Tair"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tair=28.6 C in methane, what would the Soil temperature from air temperature be if Tair were adjusted to 31.5? Provide the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "0.708", "answer_float": 0.708, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.708", "answer_float": 0.708, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.708", "answer_float": 0.708, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.708", "answer_float": 0.708, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.708", "answer_float": 0.708, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.708", "answer_float": 0.708, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.606 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "50.61", "answer_float": 50.61, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "50.61", "answer_float": 50.61, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "50.61", "answer_float": 50.61, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "50.61", "answer_float": 50.61, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "50.61", "answer_float": 50.61, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "50.61", "answer_float": 50.61, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given GrainYield=8113.0 kg/ha, NUptake=160.3 kg N/ha, calculate the Nitrogen use efficiency for a nitrogen. The result should be in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "1.04", "answer_float": 1.04, "change": {"new_value": 0.59, "old_value": 0.65, "parameter": "beta"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.04", "answer_float": 1.04, "change": {"new_value": 0.59, "old_value": 0.65, "parameter": "beta"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.04", "answer_float": 1.04, "change": {"new_value": 0.59, "old_value": 0.65, "parameter": "beta"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.04", "answer_float": 1.04, "change": {"new_value": 0.59, "old_value": 0.65, "parameter": "beta"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.04", "answer_float": 1.04, "change": {"new_value": 0.59, "old_value": 0.65, "parameter": "beta"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.04", "answer_float": 1.04, "change": {"new_value": 0.59, "old_value": 0.65, "parameter": "beta"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=363.0 ppm, beta=0.65 dimensionless. After changing beta from 0.65 to 0.59, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "183.14", "answer_float": 183.14, "change": {"new_value": 223.6, "old_value": 248.4, "parameter": "SLAc"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "183.14", "answer_float": 183.14, "change": {"new_value": 223.6, "old_value": 248.4, "parameter": "SLAc"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "183.14", "answer_float": 183.14, "change": {"new_value": 223.6, "old_value": 248.4, "parameter": "SLAc"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "183.14", "answer_float": 183.14, "change": {"new_value": 223.6, "old_value": 248.4, "parameter": "SLAc"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "183.14", "answer_float": 183.14, "change": {"new_value": 223.6, "old_value": 248.4, "parameter": "SLAc"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "183.14", "answer_float": 183.14, "change": {"new_value": 223.6, "old_value": 248.4, "parameter": "SLAc"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Specific leaf area responds to parameter changes. Baseline: GDD=611.2 C day, SLAc=248.4 cm2/g. After changing SLAc from 248.4 to 223.6, compute the new Specific leaf area. Give the numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "0.35", "answer_float": 0.35, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.35", "answer_float": 0.35, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.35", "answer_float": 0.35, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.35", "answer_float": 0.35, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.35", "answer_float": 0.35, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.35", "answer_float": 0.35, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Physiological age factor. The measured inputs are PDT=47.6 physiological days, a=0.053 dimensionless. What is the physiological age factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "56.2", "answer_float": 56.2, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "C day"} +{"answer": "56.2", "answer_float": 56.2, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "C day"} +{"answer": "56.2", "answer_float": 56.2, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "C day"} +{"answer": "56.2", "answer_float": 56.2, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "C day"} +{"answer": "56.2", "answer_float": 56.2, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "C day"} +{"answer": "56.2", "answer_float": 56.2, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Emergence thermal requirement is computed as part of phenology modeling. With SDEPTH=1.6 cm, determine the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "C day"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 44.7, "old_value": 38.9, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 44.7, "old_value": 38.9, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 44.7, "old_value": 38.9, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 44.7, "old_value": 38.9, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 44.7, "old_value": 38.9, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 44.7, "old_value": 38.9, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=38.9 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 44.7. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 3, "problem": "In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 3, "problem": "In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 3, "problem": "In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 3, "problem": "In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 3, "problem": "In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 3, "problem": "In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a photosynthesis, the Canopy extinction coefficient is computed as part of photosynthesis modeling. With PDT=2.2 physiological days, determine the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.763", "answer_float": 0.763, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.763", "answer_float": 0.763, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.763", "answer_float": 0.763, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.763", "answer_float": 0.763, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.763", "answer_float": 0.763, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.763", "answer_float": 0.763, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Methane production. The measured inputs are FEh=0.428 fraction, TI=0.722 fraction, Cr=4.31 g C/m2/day, Com=3.488 g C/m2/day. What is the methane production? Give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.993", "answer_float": 0.993, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.993", "answer_float": 0.993, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.993", "answer_float": 0.993, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.993", "answer_float": 0.993, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.993", "answer_float": 0.993, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.993", "answer_float": 0.993, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.428 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "244.63", "answer_float": 244.63, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "244.63", "answer_float": 244.63, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "244.63", "answer_float": 244.63, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "244.63", "answer_float": 244.63, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "244.63", "answer_float": 244.63, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "244.63", "answer_float": 244.63, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Daily dry matter increment is computed as part of respiration modeling. With DTGA=432.5 kg/ha/day, RM=84.6 kg/ha/day, RG=115.5 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "1.30", "answer_float": 1.3, "change": {"new_value": 0.55, "old_value": 0.5, "parameter": "beta"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.30", "answer_float": 1.3, "change": {"new_value": 0.55, "old_value": 0.5, "parameter": "beta"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.30", "answer_float": 1.3, "change": {"new_value": 0.55, "old_value": 0.5, "parameter": "beta"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.30", "answer_float": 1.3, "change": {"new_value": 0.55, "old_value": 0.5, "parameter": "beta"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.30", "answer_float": 1.3, "change": {"new_value": 0.55, "old_value": 0.5, "parameter": "beta"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配��数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.30", "answer_float": 1.3, "change": {"new_value": 0.55, "old_value": 0.5, "parameter": "beta"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how CO2 response factor responds to parameter changes. Baseline: Cx=583.0 ppm, beta=0.5 dimensionless. After changing beta from 0.5 to 0.55, compute the new CO2 response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Exponential growth", "mode": "direct", "precision": 2, "problem": "Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Exponential growth", "mode": "direct", "precision": 2, "problem": "Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Potential leaf area index (潜在叶面积指数).\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Exponential growth", "mode": "direct", "precision": 2, "problem": "Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Exponential growth", "mode": "direct", "precision": 2, "problem": "Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Exponential growth", "mode": "direct", "precision": 2, "problem": "Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Exponential growth", "mode": "direct", "precision": 2, "problem": "Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given LAI0=0.279 m2/m2, RP=0.0063 d-1, GDD=203.4 C day, calculate the Potential leaf area index for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "12.52", "answer_float": 12.52, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Power response", "mode": "direct", "precision": 2, "problem": "A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.52", "answer_float": 12.52, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Power response", "mode": "direct", "precision": 2, "problem": "A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.52", "answer_float": 12.52, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Power response", "mode": "direct", "precision": 2, "problem": "A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.52", "answer_float": 12.52, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Power response", "mode": "direct", "precision": 2, "problem": "A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.52", "answer_float": 12.52, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Power response", "mode": "direct", "precision": 2, "problem": "A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.52", "answer_float": 12.52, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Power response", "mode": "direct", "precision": 2, "problem": "A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Critical daylength. The measured inputs are Po=10.5 h, PS=0.246 dimensionless. What is the critical daylength? Give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "1.222", "answer_float": 1.222, "change": {"new_value": 551.0, "old_value": 501.0, "parameter": "CO2"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.222", "answer_float": 1.222, "change": {"new_value": 551.0, "old_value": 501.0, "parameter": "CO2"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.222", "answer_float": 1.222, "change": {"new_value": 551.0, "old_value": 501.0, "parameter": "CO2"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.222", "answer_float": 1.222, "change": {"new_value": 551.0, "old_value": 501.0, "parameter": "CO2"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.222", "answer_float": 1.222, "change": {"new_value": 551.0, "old_value": 501.0, "parameter": "CO2"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.222", "answer_float": 1.222, "change": {"new_value": 551.0, "old_value": 501.0, "parameter": "CO2"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are CO2=501.0 ppm, Coef_B_CO2=0.46 dimensionless. A field technician asks what the carbon-nitrogen cycling co2 fertilization factor would become if CO2 changed to 551.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "0.802", "answer_float": 0.802, "change": {"new_value": 33.3, "old_value": 29.0, "parameter": "Td"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_sensitivity_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "0.802", "answer_float": 0.802, "change": {"new_value": 33.3, "old_value": 29.0, "parameter": "Td"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_sensitivity_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "0.802", "answer_float": 0.802, "change": {"new_value": 33.3, "old_value": 29.0, "parameter": "Td"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "0.802", "answer_float": 0.802, "change": {"new_value": 33.3, "old_value": 29.0, "parameter": "Td"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "0.802", "answer_float": 0.802, "change": {"new_value": 33.3, "old_value": 29.0, "parameter": "Td"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "0.802", "answer_float": 0.802, "change": {"new_value": 33.3, "old_value": 29.0, "parameter": "Td"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling stress, we analyze how carbon-nitrogen cycling crop temperature activity factor responds to parameter changes. Baseline: Td=29.0 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.13 dimensionless. After changing Td from 29.0 to 33.3, compute the new carbon-nitrogen cycling crop temperature activity factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "0.265", "answer_float": 0.265, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.265", "answer_float": 0.265, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.265", "answer_float": 0.265, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.265", "answer_float": 0.265, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.265", "answer_float": 0.265, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.265", "answer_float": 0.265, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Daily physiological effectiveness. The inputs are PDT=18.9 physiological days, DTE=0.265 fraction, IE=0.85 dimensionless, RPE=0.6 fraction, BFF=1.17 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.687", "answer_float": 0.687, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.687", "answer_float": 0.687, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.687", "answer_float": 0.687, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.687", "answer_float": 0.687, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.687", "answer_float": 0.687, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.687", "answer_float": 0.687, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.617 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.329", "answer_float": 0.329, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.329", "answer_float": 0.329, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Plant methane transport factor (植株传输因子).\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.329", "answer_float": 0.329, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.329", "answer_float": 0.329, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.329", "answer_float": 0.329, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.329", "answer_float": 0.329, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Plant methane transport factor. The measured inputs are CH4RiceEfC=0.39 fraction, W=5043.0 kg/ha, Wmax=10266.0 kg/ha. What is the plant methane transport factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_direct_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Piecewise exponential", "mode": "direct", "precision": 3, "problem": "Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_direct_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise exponential", "mode": "direct", "precision": 3, "problem": "Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_direct_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise exponential", "mode": "direct", "precision": 3, "problem": "Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_direct_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise exponential", "mode": "direct", "precision": 3, "problem": "Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_direct_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise exponential", "mode": "direct", "precision": 3, "problem": "Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_direct_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise exponential", "mode": "direct", "precision": 3, "problem": "Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Eh=-204.5 mV, calculate the Redox potential factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.637", "answer_float": 0.637, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_chain_2_t1_0001__C", "is_gold": false, "level": 4, "math_type": "Double logistic", "mode": "chain_2", "precision": 3, "problem": "Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.637", "answer_float": 0.637, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_chain_2_t1_0001__K_name", "is_gold": false, "level": 4, "math_type": "Double logistic", "mode": "chain_2", "precision": 3, "problem": "Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response solar interception fraction (环境响应光截获比例).\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.637", "answer_float": 0.637, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_chain_2_t1_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Double logistic", "mode": "chain_2", "precision": 3, "problem": "Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.637", "answer_float": 0.637, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_chain_2_t1_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Double logistic", "mode": "chain_2", "precision": 3, "problem": "Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.637", "answer_float": 0.637, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_chain_2_t1_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Double logistic", "mode": "chain_2", "precision": 3, "problem": "Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.637", "answer_float": 0.637, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_chain_2_t1_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Double logistic", "mode": "chain_2", "precision": 3, "problem": "Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given fSolarMax=0.95 fraction, TT=912.5 C day, I50A=483.6 C day, I50B=435.8 C day, Tsum=2171.0 C day, fSolar_water=0.68 fraction, compute environmental-response solar interception fraction in environmental-response formula system. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.007", "answer_float": 0.007, "change": {"new_value": 36.0, "old_value": 32.3, "parameter": "t_soil"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_sensitivity_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Logarithmic minimum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.007", "answer_float": 0.007, "change": {"new_value": 36.0, "old_value": 32.3, "parameter": "t_soil"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_sensitivity_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Logarithmic minimum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.007", "answer_float": 0.007, "change": {"new_value": 36.0, "old_value": 32.3, "parameter": "t_soil"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Logarithmic minimum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.007", "answer_float": 0.007, "change": {"new_value": 36.0, "old_value": 32.3, "parameter": "t_soil"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Logarithmic minimum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.007", "answer_float": 0.007, "change": {"new_value": 36.0, "old_value": 32.3, "parameter": "t_soil"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Logarithmic minimum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.007", "answer_float": 0.007, "change": {"new_value": 36.0, "old_value": 32.3, "parameter": "t_soil"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Logarithmic minimum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane ebullition emission responds to parameter changes. Baseline: P=2.174 g CH4/m2/day, t_soil=32.3 C, Wr=735.6 kg/ha. After changing t_soil from 32.3 to 36.0, compute the new Methane ebullition emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "1.898", "answer_float": 1.898, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.898", "answer_float": 1.898, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.898", "answer_float": 1.898, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.898", "answer_float": 1.898, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.898", "answer_float": 1.898, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.898", "answer_float": 1.898, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=6.44 umol CO2/m2/s, F_T=0.73 fraction, F_Water=0.49 fraction, F_CO2=1.25 fraction, Day_Length_hour=15.26 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "195.60", "answer_float": 195.6, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "195.60", "answer_float": 195.6, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "195.60", "answer_float": 195.6, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "195.60", "answer_float": 195.6, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "195.60", "answer_float": 195.6, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "195.60", "answer_float": 195.6, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Growth respiration is computed as part of respiration modeling. With Rg=0.4 fraction, DTGA=489.0 kg/ha/day, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "20.022", "answer_float": 20.022, "change": {"new_value": 3240.2, "old_value": 3600.2, "parameter": "W"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "20.022", "answer_float": 20.022, "change": {"new_value": 3240.2, "old_value": 3600.2, "parameter": "W"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root-exudate carbon substrate (根系分泌碳底物).\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "20.022", "answer_float": 20.022, "change": {"new_value": 3240.2, "old_value": 3600.2, "parameter": "W"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "20.022", "answer_float": 20.022, "change": {"new_value": 3240.2, "old_value": 3600.2, "parameter": "W"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "20.022", "answer_float": 20.022, "change": {"new_value": 3240.2, "old_value": 3600.2, "parameter": "W"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "20.022", "answer_float": 20.022, "change": {"new_value": 3240.2, "old_value": 3600.2, "parameter": "W"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Root-exudate carbon substrate responds to parameter changes. Baseline: VI=0.7 fraction, SI=0.65 fraction, W=3600.2 kg/ha. After changing W from 3600.2 to 3240.2, compute the new Root-exudate carbon substrate. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.580", "answer_float": 0.58, "change": {"new_value": 13.5, "old_value": 11.7, "parameter": "Tmn"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.580", "answer_float": 0.58, "change": {"new_value": 13.5, "old_value": 11.7, "parameter": "Tmn"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.580", "answer_float": 0.58, "change": {"new_value": 13.5, "old_value": 11.7, "parameter": "Tmn"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.580", "answer_float": 0.58, "change": {"new_value": 13.5, "old_value": 11.7, "parameter": "Tmn"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.580", "answer_float": 0.58, "change": {"new_value": 13.5, "old_value": 11.7, "parameter": "Tmn"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.580", "answer_float": 0.58, "change": {"new_value": 13.5, "old_value": 11.7, "parameter": "Tmn"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=279.1 g C/m2, Coef_Rm=0.002 g C/g C/s, Tmn=11.7 C, DVI=0.51 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Tmn were adjusted to 13.5? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_branch_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise trigonometric", "mode": "branch", "precision": 2, "problem": "Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_branch_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise trigonometric", "mode": "branch", "precision": 2, "problem": "Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_branch_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise trigonometric", "mode": "branch", "precision": 2, "problem": "Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_branch_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise trigonometric", "mode": "branch", "precision": 2, "problem": "Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_branch_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise trigonometric", "mode": "branch", "precision": 2, "problem": "Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_branch_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise trigonometric", "mode": "branch", "precision": 2, "problem": "Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Temperature response factor in temperature response with these inputs: T=30.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF032_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_chain_2_t1_0001__C", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "chain_2", "precision": 2, "problem": "Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_chain_2_t1_0001__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "chain_2", "precision": 2, "problem": "Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative photoperiod effectiveness (相对光周期效应).\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_chain_2_t1_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "chain_2", "precision": 2, "problem": "Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_chain_2_t1_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "chain_2", "precision": 2, "problem": "Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_chain_2_t1_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "chain_2", "precision": 2, "problem": "Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_chain_2_t1_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "chain_2", "precision": 2, "problem": "Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=10.17 h, Po=10.5 h, PS=0.121 dimensionless, compute Relative photoperiod effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "0.196", "answer_float": 0.196, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t0_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2/day"} +{"answer": "0.196", "answer_float": 0.196, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t0_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2/day"} +{"answer": "0.196", "answer_float": 0.196, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t0_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2/day"} +{"answer": "0.196", "answer_float": 0.196, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t0_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2/day"} +{"answer": "0.196", "answer_float": 0.196, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t0_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2/day"} +{"answer": "0.196", "answer_float": 0.196, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t0_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation estimates carbon-nitrogen cycling crop nitrogen demand. The measured inputs are N_shoot=4.79 g N/m2, Coef_Plant_TN_uptake=0.058 d-1, N_requirement_max_day=1.44 g N/m2/day, NPP=13.98 g C/m2/day, Ratio_N_C_photo_product=0.014 g N/g C, DVI=0.29 dimensionless. What is the carbon-nitrogen cycling crop nitrogen demand? Give the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2/day"} +{"answer": "1.028", "answer_float": 1.028, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 1, "unit": "response factor"} +{"answer": "1.028", "answer_float": 1.028, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 1, "unit": "response factor"} +{"answer": "1.028", "answer_float": 1.028, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 1, "unit": "response factor"} +{"answer": "1.028", "answer_float": 1.028, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 1, "unit": "response factor"} +{"answer": "1.028", "answer_float": 1.028, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 1, "unit": "response factor"} +{"answer": "1.028", "answer_float": 1.028, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling stress, compute the carbon-nitrogen cycling crop temperature activity factor. The inputs are Td=25.7 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.52 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 1, "unit": "response factor"} +{"answer": "0.278", "answer_float": 0.278, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_direct_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise linear", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.278", "answer_float": 0.278, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_direct_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise linear", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.278", "answer_float": 0.278, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_direct_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise linear", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.278", "answer_float": 0.278, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_direct_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise linear", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.278", "answer_float": 0.278, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_direct_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise linear", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.278", "answer_float": 0.278, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_direct_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise linear", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response temperature response is computed as part of environmental-response formula calculation. With Tmean=12.5 C, Tbase=6.6 C, Topt=27.8 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.0020", "answer_float": 0.002, "change": {"new_value": 0.45, "old_value": 0.5, "parameter": "NNI"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "precision": 4, "problem": "In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0020", "answer_float": 0.002, "change": {"new_value": 0.45, "old_value": 0.5, "parameter": "NNI"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "precision": 4, "problem": "In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0020", "answer_float": 0.002, "change": {"new_value": 0.45, "old_value": 0.5, "parameter": "NNI"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "precision": 4, "problem": "In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0020", "answer_float": 0.002, "change": {"new_value": 0.45, "old_value": 0.5, "parameter": "NNI"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "precision": 4, "problem": "In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0020", "answer_float": 0.002, "change": {"new_value": 0.45, "old_value": 0.5, "parameter": "NNI"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "precision": 4, "problem": "In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0020", "answer_float": 0.002, "change": {"new_value": 0.45, "old_value": 0.5, "parameter": "NNI"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Minimum stress multiplier", "mode": "sensitivity", "precision": 4, "problem": "In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Actual LAI relative growth rate responds to parameter changes. Baseline: RP=0.0044 d-1, NNI=0.5 fraction, WDF=0.92 fraction. After changing NNI from 0.5 to 0.45, compute the new Actual LAI relative growth rate. Give the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.867", "answer_float": 0.867, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "0.867", "answer_float": 0.867, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "0.867", "answer_float": 0.867, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "0.867", "answer_float": 0.867, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "0.867", "answer_float": 0.867, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "0.867", "answer_float": 0.867, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop water stress factor in carbon-nitrogen cycling water with these inputs: Soil_Water=0.574 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "0.999", "answer_float": 0.999, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_direct_t0_0000__C", "is_gold": false, "level": 1, "math_type": "Logistic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.999", "answer_float": 0.999, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_direct_t0_0000__K_name", "is_gold": false, "level": 1, "math_type": "Logistic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子).\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.999", "answer_float": 0.999, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_direct_t0_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Logistic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.999", "answer_float": 0.999, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_direct_t0_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Logistic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.999", "answer_float": 0.999, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_direct_t0_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Logistic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF010_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.999", "answer_float": 0.999, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_direct_t0_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Logistic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling soil pH decomposition factor. The measured inputs are pH=7.85 pH. What is the carbon-nitrogen cycling soil ph decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "83.94", "answer_float": 83.94, "change": {"new_value": 299.8, "old_value": 260.7, "parameter": "DTGA"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "83.94", "answer_float": 83.94, "change": {"new_value": 299.8, "old_value": 260.7, "parameter": "DTGA"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "83.94", "answer_float": 83.94, "change": {"new_value": 299.8, "old_value": 260.7, "parameter": "DTGA"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "83.94", "answer_float": 83.94, "change": {"new_value": 299.8, "old_value": 260.7, "parameter": "DTGA"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "83.94", "answer_float": 83.94, "change": {"new_value": 299.8, "old_value": 260.7, "parameter": "DTGA"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "83.94", "answer_float": 83.94, "change": {"new_value": 299.8, "old_value": 260.7, "parameter": "DTGA"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.28 fraction, DTGA=260.7 kg/ha/day in respiration, what would the Growth respiration be if DTGA were adjusted to 299.8? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "0.960", "answer_float": 0.96, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "0.960", "answer_float": 0.96, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "0.960", "answer_float": 0.96, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "0.960", "answer_float": 0.96, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "0.960", "answer_float": 0.96, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "0.960", "answer_float": 0.96, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given T=26.9 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.35 dimensionless, calculate the Relative thermal effectiveness for a phenology. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "8.259", "answer_float": 8.259, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.259", "answer_float": 8.259, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.259", "answer_float": 8.259, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.259", "answer_float": 8.259, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.259", "answer_float": 8.259, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.259", "answer_float": 8.259, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.478 fraction, TI=0.924 fraction, Cr=61.55 g C/m2/day, Com=7.123 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.326", "answer_float": 0.326, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "0.326", "answer_float": 0.326, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "0.326", "answer_float": 0.326, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "0.326", "answer_float": 0.326, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "0.326", "answer_float": 0.326, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "0.326", "answer_float": 0.326, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane temperature index is computed as part of methane modeling. With Q10=3.52 dimensionless, t_soil=21.1 C, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "0.11", "answer_float": 0.11, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "precision": 2, "problem": "Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.11", "answer_float": 0.11, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "precision": 2, "problem": "Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.11", "answer_float": 0.11, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "precision": 2, "problem": "Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.11", "answer_float": 0.11, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "precision": 2, "problem": "Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.11", "answer_float": 0.11, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "precision": 2, "problem": "Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.11", "answer_float": 0.11, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Multi-step sigmoid", "mode": "direct", "precision": 2, "problem": "Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Y_grain=7851.0 kg/ha, B_above=17858.0 kg/ha, PDT=34.4 physiological days, calculate the Panicle partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.00", "answer_float": 0.0, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "0.00", "answer_float": 0.0, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "0.00", "answer_float": 0.0, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "0.00", "answer_float": 0.0, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "0.00", "answer_float": 0.0, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "0.00", "answer_float": 0.0, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=3.0 C, Tbase=6.8 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "29.939", "answer_float": 29.939, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_chain_2_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Michaelis-Menten response", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "29.939", "answer_float": 29.939, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_chain_2_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Michaelis-Menten response", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "29.939", "answer_float": 29.939, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_chain_2_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Michaelis-Menten response", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "29.939", "answer_float": 29.939, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_chain_2_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Michaelis-Menten response", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "29.939", "answer_float": 29.939, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Michaelis-Menten response", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "29.939", "answer_float": 29.939, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Michaelis-Menten response", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is obtained by chaining intermediate calculations. With inputs Coef_alfa=27.9 umol CO2/gN/s, Coef_beta=146.2 umol/m2/s, PAR=309.5 umol/m2/s, Leaf_N=1.58 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "0.771", "answer_float": 0.771, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.771", "answer_float": 0.771, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.771", "answer_float": 0.771, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.771", "answer_float": 0.771, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.771", "answer_float": 0.771, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.771", "answer_float": 0.771, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a phenology, the Daily physiological effectiveness is computed as part of phenology modeling. With PDT=22.1 physiological days, DTE=0.771 fraction, IE=1.24 dimensionless, RPE=0.81 fraction, BFF=1.19 dimensionless, determine the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "5.724", "answer_float": 5.724, "change": {"new_value": 0.004, "old_value": 0.0037, "parameter": "Coef_Rm"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0001__C", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "5.724", "answer_float": 5.724, "change": {"new_value": 0.004, "old_value": 0.0037, "parameter": "Coef_Rm"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0001__K_name", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "5.724", "answer_float": 5.724, "change": {"new_value": 0.004, "old_value": 0.0037, "parameter": "Coef_Rm"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "5.724", "answer_float": 5.724, "change": {"new_value": 0.004, "old_value": 0.0037, "parameter": "Coef_Rm"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "5.724", "answer_float": 5.724, "change": {"new_value": 0.004, "old_value": 0.0037, "parameter": "Coef_Rm"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "5.724", "answer_float": 5.724, "change": {"new_value": 0.004, "old_value": 0.0037, "parameter": "Coef_Rm"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Biom_plant_yesterday=941.4 g C/m2, Coef_Rm=0.0037 g C/g C/s, Tmn=19.0 C, DVI=0.78 dimensionless in carbon-nitrogen cycling respiration, what would the carbon-nitrogen cycling plant maintenance respiration be if Coef_Rm were adjusted to 0.004? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.842", "answer_float": 0.842, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_branch_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.842", "answer_float": 0.842, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_branch_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子).\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.842", "answer_float": 0.842, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_branch_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.842", "answer_float": 0.842, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_branch_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.842", "answer_float": 0.842, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_branch_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.842", "answer_float": 0.842, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_branch_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon water factor in carbon-nitrogen cycling soil carbon with these inputs: Soil_Water=0.239 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.26 g/cm3. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.978", "answer_float": 0.978, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.978", "answer_float": 0.978, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.978", "answer_float": 0.978, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.978", "answer_float": 0.978, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.978", "answer_float": 0.978, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.978", "answer_float": 0.978, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Water-content decomposition factor. The measured inputs are WaterC=0.393 fraction. What is the water-content decomposition factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "1.177", "answer_float": 1.177, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.177", "answer_float": 1.177, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.177", "answer_float": 1.177, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.177", "answer_float": 1.177, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.177", "answer_float": 1.177, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.177", "answer_float": 1.177, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling CO2 fertilization factor is computed as part of carbon-nitrogen cycling photosynthesis modeling. With CO2=459.0 ppm, Coef_B_CO2=0.59 dimensionless, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "38.77", "answer_float": 38.77, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "38.77", "answer_float": 38.77, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "38.77", "answer_float": 38.77, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "38.77", "answer_float": 38.77, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "38.77", "answer_float": 38.77, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "38.77", "answer_float": 38.77, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis module estimates Actual maximum assimilation rate using intermediate drivers. Use the provided inputs in sequence. The measurements are Am=41.5 kg CO2/ha/h, FCO2=1.05 fraction, FPA=0.93 fraction, FT=0.92 fraction, NNI=1.04 fraction, WDF=1.08 fraction. Compute the actual maximum assimilation rate and give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "5978.8", "answer_float": 5978.8, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha"} +{"answer": "5978.8", "answer_float": 5978.8, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha"} +{"answer": "5978.8", "answer_float": 5978.8, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha"} +{"answer": "5978.8", "answer_float": 5978.8, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha"} +{"answer": "5978.8", "answer_float": 5978.8, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha"} +{"answer": "5978.8", "answer_float": 5978.8, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response yield. The measured inputs are Biomass=16159.0 kg/ha, HI=0.37 fraction. What is the simple yield? Give the final numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha"} +{"answer": "8.285", "answer_float": 8.285, "change": {"new_value": 753.4, "old_value": 684.9, "parameter": "Biom_shoot"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF005", "formula_name_en": "carbon-nitrogen cycling shoot nitrogen allocation", "id": "CMF005_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF005_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2"} +{"answer": "8.285", "answer_float": 8.285, "change": {"new_value": 753.4, "old_value": 684.9, "parameter": "Biom_shoot"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF005", "formula_name_en": "carbon-nitrogen cycling shoot nitrogen allocation", "id": "CMF005_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量).\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF005_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2"} +{"answer": "8.285", "answer_float": 8.285, "change": {"new_value": 753.4, "old_value": 684.9, "parameter": "Biom_shoot"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF005", "formula_name_en": "carbon-nitrogen cycling shoot nitrogen allocation", "id": "CMF005_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF005_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2"} +{"answer": "8.285", "answer_float": 8.285, "change": {"new_value": 753.4, "old_value": 684.9, "parameter": "Biom_shoot"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF005", "formula_name_en": "carbon-nitrogen cycling shoot nitrogen allocation", "id": "CMF005_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF005_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2"} +{"answer": "8.285", "answer_float": 8.285, "change": {"new_value": 753.4, "old_value": 684.9, "parameter": "Biom_shoot"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF005", "formula_name_en": "carbon-nitrogen cycling shoot nitrogen allocation", "id": "CMF005_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF005_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2"} +{"answer": "8.285", "answer_float": 8.285, "change": {"new_value": 753.4, "old_value": 684.9, "parameter": "Biom_shoot"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF005", "formula_name_en": "carbon-nitrogen cycling shoot nitrogen allocation", "id": "CMF005_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Biomass-weighted allocation", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling nitrogen calculation involves sensitivity analysis. The current measurements are Plant_TN_uptake=8.55 g N/m2, Biom_shoot=684.9 g C/m2, Biom_root=24.1 g C/m2. A field technician asks what the carbon-nitrogen cycling shoot nitrogen allocation would become if Biom_shoot changed to 753.4. Report the new final numeric value in g N/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF005_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 0, "unit": "g N/m2"} +{"answer": "439.26", "answer_float": 439.26, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "439.26", "answer_float": 439.26, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "439.26", "answer_float": 439.26, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "439.26", "answer_float": 439.26, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "439.26", "answer_float": 439.26, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "439.26", "answer_float": 439.26, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=565.5 kg/ha/day, RM=94.5 kg/ha/day, RG=53.7 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "12.90", "answer_float": 12.9, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "12.90", "answer_float": 12.9, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "12.90", "answer_float": 12.9, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "12.90", "answer_float": 12.9, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "12.90", "answer_float": 12.9, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "12.90", "answer_float": 12.9, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_branch_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Maximum linear", "mode": "branch", "precision": 2, "problem": "Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response daily thermal time in environmental-response formula system with these inputs: Tmean=23.5 C, Tbase=10.6 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "C day"} +{"answer": "0.331", "answer_float": 0.331, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.331", "answer_float": 0.331, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.331", "answer_float": 0.331, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.331", "answer_float": 0.331, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.331", "answer_float": 0.331, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.331", "answer_float": 0.331, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Soil organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.275 fraction, SI=1.15 fraction, TI=0.28 fraction, OMS=1246.5 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "6.319", "answer_float": 6.319, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_direct_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Chained sum", "mode": "direct", "precision": 3, "problem": "In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.319", "answer_float": 6.319, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_direct_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Chained sum", "mode": "direct", "precision": 3, "problem": "In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.319", "answer_float": 6.319, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_direct_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Chained sum", "mode": "direct", "precision": 3, "problem": "In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.319", "answer_float": 6.319, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_direct_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Chained sum", "mode": "direct", "precision": 3, "problem": "In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.319", "answer_float": 6.319, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_direct_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Chained sum", "mode": "direct", "precision": 3, "problem": "In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.319", "answer_float": 6.319, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_direct_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Chained sum", "mode": "direct", "precision": 3, "problem": "In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Total methane emission is computed as part of methane modeling. With Ebl=4.391 g CH4/m2/day, P=11.616 g CH4/m2/day, CH4RiceEfC=0.208 fraction, W=5453.7 kg/ha, Wmax=9178.0 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.345", "answer_float": 0.345, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0001__C", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.345", "answer_float": 0.345, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0001__K_name", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.345", "answer_float": 0.345, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.345", "answer_float": 0.345, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.345", "answer_float": 0.345, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.345", "answer_float": 0.345, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=34.5 physiological days, DTE=0.515 fraction, IE=1.24 dimensionless, RPE=0.49 fraction, BFF=0.67 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "27.20", "answer_float": 27.2, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0000__C", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "27.20", "answer_float": 27.2, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0000__K_name", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "27.20", "answer_float": 27.2, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "27.20", "answer_float": 27.2, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "27.20", "answer_float": 27.2, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "27.20", "answer_float": 27.2, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=30.0 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "0.700", "answer_float": 0.7, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.700", "answer_float": 0.7, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.700", "answer_float": 0.7, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.700", "answer_float": 0.7, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.700", "answer_float": 0.7, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.700", "answer_float": 0.7, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Soil_Water=0.178 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, calculate the carbon-nitrogen cycling crop water stress factor for a carbon-nitrogen cycling water. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "151.50", "answer_float": 151.5, "change": {"new_value": 0.41, "old_value": 0.36, "parameter": "Rg"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "151.50", "answer_float": 151.5, "change": {"new_value": 0.41, "old_value": 0.36, "parameter": "Rg"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "151.50", "answer_float": 151.5, "change": {"new_value": 0.41, "old_value": 0.36, "parameter": "Rg"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "151.50", "answer_float": 151.5, "change": {"new_value": 0.41, "old_value": 0.36, "parameter": "Rg"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "151.50", "answer_float": 151.5, "change": {"new_value": 0.41, "old_value": 0.36, "parameter": "Rg"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "151.50", "answer_float": 151.5, "change": {"new_value": 0.41, "old_value": 0.36, "parameter": "Rg"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Rg=0.36 fraction, DTGA=369.5 kg/ha/day in respiration, what would the Growth respiration be if Rg were adjusted to 0.41? Provide the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "0.475", "answer_float": 0.475, "change": {"new_value": 1.22, "old_value": 1.35, "parameter": "ARID"}, "condition": "C", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "response factor"} +{"answer": "0.475", "answer_float": 0.475, "change": {"new_value": 1.22, "old_value": 1.35, "parameter": "ARID"}, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response water stress response (环境响应水分响应).\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "response factor"} +{"answer": "0.475", "answer_float": 0.475, "change": {"new_value": 1.22, "old_value": 1.35, "parameter": "ARID"}, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "response factor"} +{"answer": "0.475", "answer_float": 0.475, "change": {"new_value": 1.22, "old_value": 1.35, "parameter": "ARID"}, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "response factor"} +{"answer": "0.475", "answer_float": 0.475, "change": {"new_value": 1.22, "old_value": 1.35, "parameter": "ARID"}, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "response factor"} +{"answer": "0.475", "answer_float": 0.475, "change": {"new_value": 1.22, "old_value": 1.35, "parameter": "ARID"}, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements ARID=1.35 dimensionless, S_Water=0.43 dimensionless in environmental-response formula system, what would the environmental-response water stress response be if ARID were adjusted to 1.22? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "response factor"} +{"answer": "0.08", "answer_float": 0.08, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.08", "answer_float": 0.08, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.08", "answer_float": 0.08, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.08", "answer_float": 0.08, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.08", "answer_float": 0.08, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.08", "answer_float": 0.08, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=51.0 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.0077", "answer_float": 0.0077, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0077", "answer_float": 0.0077, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0077", "answer_float": 0.0077, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0077", "answer_float": 0.0077, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0077", "answer_float": 0.0077, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0077", "answer_float": 0.0077, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=14.5 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.34", "answer_float": 0.34, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_branch_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise exponential decay", "mode": "branch", "precision": 2, "problem": "Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "0.34", "answer_float": 0.34, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_branch_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise exponential decay", "mode": "branch", "precision": 2, "problem": "Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "0.34", "answer_float": 0.34, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_branch_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise exponential decay", "mode": "branch", "precision": 2, "problem": "Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "0.34", "answer_float": 0.34, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_branch_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise exponential decay", "mode": "branch", "precision": 2, "problem": "Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "0.34", "answer_float": 0.34, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_branch_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise exponential decay", "mode": "branch", "precision": 2, "problem": "Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "0.34", "answer_float": 0.34, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_branch_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise exponential decay", "mode": "branch", "precision": 2, "problem": "Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Physiological age factor in photosynthesis with these inputs: PDT=50.0 physiological days, a=0.049 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "14.04", "answer_float": 14.04, "change": {"new_value": 0.08, "old_value": 0.083, "parameter": "PS"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "14.04", "answer_float": 14.04, "change": {"new_value": 0.08, "old_value": 0.083, "parameter": "PS"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "14.04", "answer_float": 14.04, "change": {"new_value": 0.08, "old_value": 0.083, "parameter": "PS"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "14.04", "answer_float": 14.04, "change": {"new_value": 0.08, "old_value": 0.083, "parameter": "PS"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "14.04", "answer_float": 14.04, "change": {"new_value": 0.08, "old_value": 0.083, "parameter": "PS"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "14.04", "answer_float": 14.04, "change": {"new_value": 0.08, "old_value": 0.083, "parameter": "PS"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, we analyze how Critical daylength responds to parameter changes. Baseline: Po=10.5 h, PS=0.083 dimensionless. After changing PS from 0.083 to 0.08, compute the new Critical daylength. Give the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_branch_t1_0001__C", "is_gold": false, "level": 3, "math_type": "Q10 with temperature clipping", "mode": "branch", "precision": 3, "problem": "For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_branch_t1_0001__K_name", "is_gold": false, "level": 3, "math_type": "Q10 with temperature clipping", "mode": "branch", "precision": 3, "problem": "For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_branch_t1_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Q10 with temperature clipping", "mode": "branch", "precision": 3, "problem": "For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_branch_t1_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Q10 with temperature clipping", "mode": "branch", "precision": 3, "problem": "For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_branch_t1_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Q10 with temperature clipping", "mode": "branch", "precision": 3, "problem": "For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_branch_t1_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Q10 with temperature clipping", "mode": "branch", "precision": 3, "problem": "For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For methane, compute the Methane temperature index. The inputs are Q10=2.2 dimensionless, t_soil=39.9 C. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "1.18", "answer_float": 1.18, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 2, "problem": "Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.18", "answer_float": 1.18, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 2, "problem": "Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.18", "answer_float": 1.18, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 2, "problem": "Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.18", "answer_float": 1.18, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 2, "problem": "Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.18", "answer_float": 1.18, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 2, "problem": "Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.18", "answer_float": 1.18, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 2, "problem": "Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Cx=419.0 ppm, beta=0.86 dimensionless, calculate the CO2 response factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "0.198", "answer_float": 0.198, "change": {"new_value": -7.0, "old_value": -6.1, "parameter": "Eh"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t0_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.198", "answer_float": 0.198, "change": {"new_value": -7.0, "old_value": -6.1, "parameter": "Eh"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t0_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.198", "answer_float": 0.198, "change": {"new_value": -7.0, "old_value": -6.1, "parameter": "Eh"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.198", "answer_float": 0.198, "change": {"new_value": -7.0, "old_value": -6.1, "parameter": "Eh"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.198", "answer_float": 0.198, "change": {"new_value": -7.0, "old_value": -6.1, "parameter": "Eh"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.198", "answer_float": 0.198, "change": {"new_value": -7.0, "old_value": -6.1, "parameter": "Eh"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Eh=-6.1 mV. A field technician asks what the redox potential factor would become if Eh changed to -7.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "71.6", "answer_float": 71.6, "change": {"new_value": 3.8, "old_value": 4.2, "parameter": "SDEPTH"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 1, "problem": "Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "C day"} +{"answer": "71.6", "answer_float": 71.6, "change": {"new_value": 3.8, "old_value": 4.2, "parameter": "SDEPTH"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 1, "problem": "Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "C day"} +{"answer": "71.6", "answer_float": 71.6, "change": {"new_value": 3.8, "old_value": 4.2, "parameter": "SDEPTH"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 1, "problem": "Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "C day"} +{"answer": "71.6", "answer_float": 71.6, "change": {"new_value": 3.8, "old_value": 4.2, "parameter": "SDEPTH"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 1, "problem": "Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "C day"} +{"answer": "71.6", "answer_float": 71.6, "change": {"new_value": 3.8, "old_value": 4.2, "parameter": "SDEPTH"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 1, "problem": "Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "C day"} +{"answer": "71.6", "answer_float": 71.6, "change": {"new_value": 3.8, "old_value": 4.2, "parameter": "SDEPTH"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 1, "problem": "Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements SDEPTH=4.2 cm in phenology, what would the Emergence thermal requirement be if SDEPTH were adjusted to 3.8? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "C day"} +{"answer": "10.69", "answer_float": 10.69, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "10.69", "answer_float": 10.69, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "10.69", "answer_float": 10.69, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "10.69", "answer_float": 10.69, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "10.69", "answer_float": 10.69, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "10.69", "answer_float": 10.69, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Actual maximum assimilation rate. The measured inputs are Am=28.3 kg CO2/ha/h, FCO2=1.13 fraction, FPA=0.91 fraction, FT=0.54 fraction, NNI=0.68 fraction, WDF=0.73 fraction. What is the actual maximum assimilation rate? Give the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "0.962", "answer_float": 0.962, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.962", "answer_float": 0.962, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.962", "answer_float": 0.962, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.962", "answer_float": 0.962, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.962", "answer_float": 0.962, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产���)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.962", "answer_float": 0.962, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Daily physiological effectiveness using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=48.6 physiological days, DTE=0.925 fraction, IE=0.89 dimensionless, RPE=0.52 fraction, BFF=1.04 dimensionless. Compute the daily physiological effectiveness and give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.07", "answer_float": 0.07, "change": {"new_value": 57.0, "old_value": 56.5, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.07", "answer_float": 0.07, "change": {"new_value": 57.0, "old_value": 56.5, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.07", "answer_float": 0.07, "change": {"new_value": 57.0, "old_value": 56.5, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.07", "answer_float": 0.07, "change": {"new_value": 57.0, "old_value": 56.5, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.07", "answer_float": 0.07, "change": {"new_value": 57.0, "old_value": 56.5, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.07", "answer_float": 0.07, "change": {"new_value": 57.0, "old_value": 56.5, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=56.5 physiological days. A field technician asks what the root partitioning index would become if PDT changed to 57.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.982", "answer_float": 0.982, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_branch_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.982", "answer_float": 0.982, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_branch_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.982", "answer_float": 0.982, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_branch_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.982", "answer_float": 0.982, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_branch_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.982", "answer_float": 0.982, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_branch_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.982", "answer_float": 0.982, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_branch_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are T=27.7 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.07 dimensionless. Determine the relative thermal effectiveness and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.045", "answer_float": 0.045, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_direct_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Chained multiplicative decomposition", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.045", "answer_float": 0.045, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_direct_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Chained multiplicative decomposition", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.045", "answer_float": 0.045, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_direct_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Chained multiplicative decomposition", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.045", "answer_float": 0.045, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_direct_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Chained multiplicative decomposition", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.045", "answer_float": 0.045, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_direct_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Chained multiplicative decomposition", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.045", "answer_float": 0.045, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_direct_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Chained multiplicative decomposition", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation estimates carbon-nitrogen cycling carbon pool respiration loss. The measured inputs are Pool=281.7 g C/m2, Coef_K_Decom=0.0057 d-1, F_temp=1.373 fraction, F_water=0.266 fraction, F_texture=0.899 fraction, F_pH=0.101 fraction, F_notill=0.93 fraction, Coef_K_Trans=0.086 fraction. What is the carbon-nitrogen cycling carbon pool respiration loss? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.17", "answer_float": 0.17, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_branch_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise logistic", "mode": "branch", "precision": 2, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_branch_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise logistic", "mode": "branch", "precision": 2, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_branch_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise logistic", "mode": "branch", "precision": 2, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_branch_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise logistic", "mode": "branch", "precision": 2, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_branch_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise logistic", "mode": "branch", "precision": 2, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_branch_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise logistic", "mode": "branch", "precision": 2, "problem": "A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=33.5 physiological days, Tmax_day=38.5 C. Determine the high-temperature fertility factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": {"new_value": 34.8, "old_value": 38.7, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_sensitivity_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": {"new_value": 34.8, "old_value": 38.7, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_sensitivity_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": {"new_value": 34.8, "old_value": 38.7, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": {"new_value": 34.8, "old_value": 38.7, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": {"new_value": 34.8, "old_value": 38.7, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.17", "answer_float": 0.17, "change": {"new_value": 34.8, "old_value": 38.7, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise logistic", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=38.7 physiological days, Tmax_day=38.5 C in temperature stress, what would the High-temperature fertility factor be if PDT were adjusted to 34.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "14.11", "answer_float": 14.11, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_chain_2_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Multiplicative minimum stress", "mode": "chain_2", "precision": 2, "problem": "In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "14.11", "answer_float": 14.11, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_chain_2_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative minimum stress", "mode": "chain_2", "precision": 2, "problem": "In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "14.11", "answer_float": 14.11, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_chain_2_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative minimum stress", "mode": "chain_2", "precision": 2, "problem": "In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "14.11", "answer_float": 14.11, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_chain_2_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative minimum stress", "mode": "chain_2", "precision": 2, "problem": "In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "14.11", "answer_float": 14.11, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative minimum stress", "mode": "chain_2", "precision": 2, "problem": "In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "14.11", "answer_float": 14.11, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative minimum stress", "mode": "chain_2", "precision": 2, "problem": "In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, the environmental-response daily biomass increment is obtained by chaining intermediate calculations. With inputs RUE=1.44 g/MJ, fSolar=0.16 fraction, SRAD=12.72 MJ/m2/day, F_CO2=1.33 fraction, F_Temp=0.77 fraction, F_Water=0.47 fraction, F_Heat=0.84 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "0.58", "answer_float": 0.58, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation estimates Stem partitioning index. The measured inputs are PDT=26.0 physiological days, Y_grain=6715.0 kg/ha, B_above=12702.0 kg/ha. What is the stem partitioning index? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.327", "answer_float": 0.327, "change": {"new_value": 1.0, "old_value": 1.11, "parameter": "SI"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.327", "answer_float": 0.327, "change": {"new_value": 1.0, "old_value": 1.11, "parameter": "SI"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.327", "answer_float": 0.327, "change": {"new_value": 1.0, "old_value": 1.11, "parameter": "SI"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.327", "answer_float": 0.327, "change": {"new_value": 1.0, "old_value": 1.11, "parameter": "SI"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.327", "answer_float": 0.327, "change": {"new_value": 1.0, "old_value": 1.11, "parameter": "SI"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.327", "answer_float": 0.327, "change": {"new_value": 1.0, "old_value": 1.11, "parameter": "SI"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.695 fraction, SI=1.11 fraction, TI=0.749 fraction, OMN=23.3 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if SI were adjusted to 1.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "14.86", "answer_float": 14.86, "change": {"new_value": 0.95, "old_value": 1.0, "parameter": "FCO2"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_sensitivity_t0_0001__C", "is_gold": false, "level": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "14.86", "answer_float": 14.86, "change": {"new_value": 0.95, "old_value": 1.0, "parameter": "FCO2"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_sensitivity_t0_0001__K_name", "is_gold": false, "level": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "14.86", "answer_float": 14.86, "change": {"new_value": 0.95, "old_value": 1.0, "parameter": "FCO2"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "14.86", "answer_float": 14.86, "change": {"new_value": 0.95, "old_value": 1.0, "parameter": "FCO2"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "14.86", "answer_float": 14.86, "change": {"new_value": 0.95, "old_value": 1.0, "parameter": "FCO2"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "14.86", "answer_float": 14.86, "change": {"new_value": 0.95, "old_value": 1.0, "parameter": "FCO2"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Multiplicative with minimum stress", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Am=49.4 kg CO2/ha/h, FCO2=1.0 fraction, FPA=0.69 fraction, FT=0.85 fraction, NNI=0.54 fraction, WDF=0.66 fraction. A field technician asks what the actual maximum assimilation rate would become if FCO2 changed to 0.95. Report the new final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "kg CO2/ha/h"} +{"answer": "8.687", "answer_float": 8.687, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "8.687", "answer_float": 8.687, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "8.687", "answer_float": 8.687, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "8.687", "answer_float": 8.687, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "8.687", "answer_float": 8.687, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "8.687", "answer_float": 8.687, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling gross primary production. The measured inputs are Photo_Rate_opt=51.04 umol CO2/m2/s, F_T=0.39 fraction, F_Water=0.66 fraction, F_CO2=0.99 fraction, Day_Length_hour=15.46 h. What is the carbon-nitrogen cycling gross primary production? Give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "6835.8", "answer_float": 6835.8, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "kg/ha"} +{"answer": "6835.8", "answer_float": 6835.8, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "kg/ha"} +{"answer": "6835.8", "answer_float": 6835.8, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "kg/ha"} +{"answer": "6835.8", "answer_float": 6835.8, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "kg/ha"} +{"answer": "6835.8", "answer_float": 6835.8, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "kg/ha"} +{"answer": "6835.8", "answer_float": 6835.8, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biomass=12428.7 kg/ha, HI=0.55 fraction, calculate the environmental-response yield for an environmental-response formula system. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "kg/ha"} +{"answer": "0.904", "answer_float": 0.904, "change": {"new_value": 0.63, "old_value": 0.55, "parameter": "RPE"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t0_0000__C", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.904", "answer_float": 0.904, "change": {"new_value": 0.63, "old_value": 0.55, "parameter": "RPE"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t0_0000__K_name", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.904", "answer_float": 0.904, "change": {"new_value": 0.63, "old_value": 0.55, "parameter": "RPE"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.904", "answer_float": 0.904, "change": {"new_value": 0.63, "old_value": 0.55, "parameter": "RPE"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.904", "answer_float": 0.904, "change": {"new_value": 0.63, "old_value": 0.55, "parameter": "RPE"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.904", "answer_float": 0.904, "change": {"new_value": 0.63, "old_value": 0.55, "parameter": "RPE"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are PDT=26.3 physiological days, DTE=0.904 fraction, IE=1.23 dimensionless, RPE=0.55 fraction, BFF=0.78 dimensionless. A field technician asks what the daily physiological effectiveness would become if RPE changed to 0.63. Report the new final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.500", "answer_float": 0.5, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.500", "answer_float": 0.5, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.500", "answer_float": 0.5, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.500", "answer_float": 0.5, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.500", "answer_float": 0.5, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.500", "answer_float": 0.5, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Pool=1787.0 g C/m2, Coef_K_Decom=0.0026 d-1, F_temp=0.649 fraction, F_water=0.566 fraction, F_texture=0.875 fraction, F_pH=0.43 fraction, F_notill=0.97 fraction, Coef_K_Trans=0.197 fraction, compute carbon-nitrogen cycling carbon pool respiration loss in carbon-nitrogen cycling soil carbon. This requires sequential calculation through intermediate variables. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.269", "answer_float": 0.269, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.269", "answer_float": 0.269, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.269", "answer_float": 0.269, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.269", "answer_float": 0.269, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.269", "answer_float": 0.269, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.269", "answer_float": 0.269, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=40.7 C, MaxT=33.1 C, ExtremeT=43.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.37", "answer_float": 0.37, "change": {"new_value": 0.044, "old_value": 0.049, "parameter": "a"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t0_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.37", "answer_float": 0.37, "change": {"new_value": 0.044, "old_value": 0.049, "parameter": "a"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t0_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.37", "answer_float": 0.37, "change": {"new_value": 0.044, "old_value": 0.049, "parameter": "a"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.37", "answer_float": 0.37, "change": {"new_value": 0.044, "old_value": 0.049, "parameter": "a"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.37", "answer_float": 0.37, "change": {"new_value": 0.044, "old_value": 0.049, "parameter": "a"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.37", "answer_float": 0.37, "change": {"new_value": 0.044, "old_value": 0.049, "parameter": "a"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=50.7 physiological days, a=0.049 dimensionless. A field technician asks what the physiological age factor would become if a changed to 0.044. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "fraction"} +{"answer": "0.57", "answer_float": 0.57, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_chain_2_t0_0001__C", "is_gold": false, "level": 4, "math_type": "Residual with piecewise components", "mode": "chain_2", "precision": 2, "problem": "A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.57", "answer_float": 0.57, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_chain_2_t0_0001__K_name", "is_gold": false, "level": 4, "math_type": "Residual with piecewise components", "mode": "chain_2", "precision": 2, "problem": "A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.57", "answer_float": 0.57, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_chain_2_t0_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Residual with piecewise components", "mode": "chain_2", "precision": 2, "problem": "A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.57", "answer_float": 0.57, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_chain_2_t0_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Residual with piecewise components", "mode": "chain_2", "precision": 2, "problem": "A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.57", "answer_float": 0.57, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_chain_2_t0_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Residual with piecewise components", "mode": "chain_2", "precision": 2, "problem": "A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.57", "answer_float": 0.57, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_chain_2_t0_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Residual with piecewise components", "mode": "chain_2", "precision": 2, "problem": "A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning module estimates Stem partitioning index using intermediate drivers. Use the provided inputs in sequence. The measurements are PDT=40.8 physiological days, Y_grain=6146.0 kg/ha, B_above=18531.0 kg/ha. Compute the stem partitioning index and give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "128.56", "answer_float": 128.56, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t0_0000__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "128.56", "answer_float": 128.56, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t0_0000__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "128.56", "answer_float": 128.56, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t0_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "128.56", "answer_float": 128.56, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t0_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "128.56", "answer_float": 128.56, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t0_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "128.56", "answer_float": 128.56, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t0_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation estimates Growth respiration. The measured inputs are Rg=0.4 fraction, DTGA=321.4 kg/ha/day. What is the growth respiration? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "4114.4", "answer_float": 4114.4, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF054", "formula_name_en": "Dry grain yield", "id": "CMF054_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF054_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Yield formation", "template_id": 1, "unit": "kg/ha"} +{"answer": "4114.4", "answer_float": 4114.4, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF054", "formula_name_en": "Dry grain yield", "id": "CMF054_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Dry grain yield (稻谷干重产量).\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF054_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Yield formation", "template_id": 1, "unit": "kg/ha"} +{"answer": "4114.4", "answer_float": 4114.4, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF054", "formula_name_en": "Dry grain yield", "id": "CMF054_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF054_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Yield formation", "template_id": 1, "unit": "kg/ha"} +{"answer": "4114.4", "answer_float": 4114.4, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF054", "formula_name_en": "Dry grain yield", "id": "CMF054_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF054_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Yield formation", "template_id": 1, "unit": "kg/ha"} +{"answer": "4114.4", "answer_float": 4114.4, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF054", "formula_name_en": "Dry grain yield", "id": "CMF054_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF054_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Yield formation", "template_id": 1, "unit": "kg/ha"} +{"answer": "4114.4", "answer_float": 4114.4, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF054", "formula_name_en": "Dry grain yield", "id": "CMF054_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given AWSP=4622.9 kg/ha, grain_fraction=0.89 fraction, calculate the Dry grain yield for a yield formation. The result should be in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF054_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Yield formation", "template_id": 1, "unit": "kg/ha"} +{"answer": "0.915", "answer_float": 0.915, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "0.915", "answer_float": 0.915, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "0.915", "answer_float": 0.915, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "0.915", "answer_float": 0.915, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "0.915", "answer_float": 0.915, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "0.915", "answer_float": 0.915, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Q10 with temperature clipping", "mode": "direct", "precision": 3, "problem": "Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Q10=3.04 dimensionless, t_soil=29.2 C, calculate the Methane temperature index for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "multiplier"} +{"answer": "3886.3", "answer_float": 3886.3, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha"} +{"answer": "3886.3", "answer_float": 3886.3, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response yield (环境响应产量).\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha"} +{"answer": "3886.3", "answer_float": 3886.3, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha"} +{"answer": "3886.3", "answer_float": 3886.3, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha"} +{"answer": "3886.3", "answer_float": 3886.3, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha"} +{"answer": "3886.3", "answer_float": 3886.3, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF062", "formula_name_en": "environmental-response yield", "id": "CMF062_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 1, "problem": "In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response yield is computed as part of environmental-response formula calculation. With Biomass=14393.7 kg/ha, HI=0.27 fraction, determine the numeric value in kg/ha.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF062_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 28.7, "old_value": 25.0, "parameter": "Tmax"}, "condition": "C", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 28.7, "old_value": 25.0, "parameter": "Tmax"}, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 28.7, "old_value": 25.0, "parameter": "Tmax"}, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 28.7, "old_value": 25.0, "parameter": "Tmax"}, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 28.7, "old_value": 25.0, "parameter": "Tmax"}, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 28.7, "old_value": 25.0, "parameter": "Tmax"}, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=25.0 C, MaxT=34.3 C, ExtremeT=47.1 C. After changing Tmax from 25.0 to 28.7, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.395", "answer_float": 0.395, "change": {"new_value": 22.6, "old_value": 20.5, "parameter": "t_soil"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "multiplier"} +{"answer": "0.395", "answer_float": 0.395, "change": {"new_value": 22.6, "old_value": 20.5, "parameter": "t_soil"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "multiplier"} +{"answer": "0.395", "answer_float": 0.395, "change": {"new_value": 22.6, "old_value": 20.5, "parameter": "t_soil"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "multiplier"} +{"answer": "0.395", "answer_float": 0.395, "change": {"new_value": 22.6, "old_value": 20.5, "parameter": "t_soil"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "multiplier"} +{"answer": "0.395", "answer_float": 0.395, "change": {"new_value": 22.6, "old_value": 20.5, "parameter": "t_soil"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "multiplier"} +{"answer": "0.395", "answer_float": 0.395, "change": {"new_value": 22.6, "old_value": 20.5, "parameter": "t_soil"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are Q10=3.51 dimensionless, t_soil=20.5 C. A field technician asks what the methane temperature index would become if t_soil changed to 22.6. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "multiplier"} +{"answer": "3.432", "answer_float": 3.432, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0001__C", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "3.432", "answer_float": 3.432, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0001__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "3.432", "answer_float": 3.432, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "3.432", "answer_float": 3.432, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "3.432", "answer_float": 3.432, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "3.432", "answer_float": 3.432, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=37.01 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.33 fraction, F_CO2=1.28 fraction, Day_Length_hour=10.37 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "17.02", "answer_float": 17.02, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t1_0001__C", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "17.02", "answer_float": 17.02, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t1_0001__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "17.02", "answer_float": 17.02, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t1_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "17.02", "answer_float": 17.02, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t1_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "17.02", "answer_float": 17.02, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t1_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "17.02", "answer_float": 17.02, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t1_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=31.1 kg CO2/ha/h, FCO2=1.2 fraction, FPA=0.92 fraction, FT=0.67 fraction, NNI=0.74 fraction, WDF=0.92 fraction, compute Actual maximum assimilation rate in photosynthesis. This requires sequential calculation through intermediate variables. Report the final numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "1.310", "answer_float": 1.31, "change": {"new_value": 0.68, "old_value": 0.75, "parameter": "Coef_B_CO2"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.310", "answer_float": 1.31, "change": {"new_value": 0.68, "old_value": 0.75, "parameter": "Coef_B_CO2"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.310", "answer_float": 1.31, "change": {"new_value": 0.68, "old_value": 0.75, "parameter": "Coef_B_CO2"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.310", "answer_float": 1.31, "change": {"new_value": 0.68, "old_value": 0.75, "parameter": "Coef_B_CO2"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.310", "answer_float": 1.31, "change": {"new_value": 0.68, "old_value": 0.75, "parameter": "Coef_B_CO2"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "1.310", "answer_float": 1.31, "change": {"new_value": 0.68, "old_value": 0.75, "parameter": "Coef_B_CO2"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling CO2 fertilization factor responds to parameter changes. Baseline: CO2=536.0 ppm, Coef_B_CO2=0.75 dimensionless. After changing Coef_B_CO2 from 0.75 to 0.68, compute the new carbon-nitrogen cycling CO2 fertilization factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "multiplier"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For temperature stress, compute the Low-temperature fertility factor. The inputs are PDT=28.5 physiological days, Tmean=16.2 C, Qt=6.2 C day. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.900", "answer_float": 0.9, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 0, "unit": "response factor"} +{"answer": "0.900", "answer_float": 0.9, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 0, "unit": "response factor"} +{"answer": "0.900", "answer_float": 0.9, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 0, "unit": "response factor"} +{"answer": "0.900", "answer_float": 0.9, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 0, "unit": "response factor"} +{"answer": "0.900", "answer_float": 0.9, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 0, "unit": "response factor"} +{"answer": "0.900", "answer_float": 0.9, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling stress calculation estimates carbon-nitrogen cycling crop temperature activity factor. The measured inputs are Td=19.3 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.16 dimensionless. What is the carbon-nitrogen cycling crop temperature activity factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 0, "unit": "response factor"} +{"answer": "0.95", "answer_float": 0.95, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_branch_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise power penalty", "mode": "branch", "precision": 2, "problem": "Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Low-temperature fertility factor in temperature stress with these inputs: PDT=30.8 physiological days, Tmean=20.6 C, Qt=4.3 C day. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "3.943", "answer_float": 3.943, "change": {"new_value": 10.12, "old_value": 11.24, "parameter": "Day_Length_hour"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "3.943", "answer_float": 3.943, "change": {"new_value": 10.12, "old_value": 11.24, "parameter": "Day_Length_hour"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "3.943", "answer_float": 3.943, "change": {"new_value": 10.12, "old_value": 11.24, "parameter": "Day_Length_hour"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "3.943", "answer_float": 3.943, "change": {"new_value": 10.12, "old_value": 11.24, "parameter": "Day_Length_hour"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "3.943", "answer_float": 3.943, "change": {"new_value": 10.12, "old_value": 11.24, "parameter": "Day_Length_hour"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "3.943", "answer_float": 3.943, "change": {"new_value": 10.12, "old_value": 11.24, "parameter": "Day_Length_hour"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=27.85 umol CO2/m2/s, F_T=0.64 fraction, F_Water=0.44 fraction, F_CO2=1.15 fraction, Day_Length_hour=11.24 h. After changing Day_Length_hour from 11.24 to 10.12, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.95", "answer_float": 0.95, "change": {"new_value": 1.6, "old_value": 1.8, "parameter": "Qt"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": {"new_value": 1.6, "old_value": 1.8, "parameter": "Qt"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": {"new_value": 1.6, "old_value": 1.8, "parameter": "Qt"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": {"new_value": 1.6, "old_value": 1.8, "parameter": "Qt"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": {"new_value": 1.6, "old_value": 1.8, "parameter": "Qt"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.95", "answer_float": 0.95, "change": {"new_value": 1.6, "old_value": 1.8, "parameter": "Qt"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=35.6 physiological days, Tmean=18.5 C, Qt=1.8 C day in temperature stress, what would the Low-temperature fertility factor be if Qt were adjusted to 1.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.166", "answer_float": 0.166, "change": {"new_value": 2200.0, "old_value": 2121.5, "parameter": "Pool"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_sensitivity_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Chained multiplicative decomposition", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.166", "answer_float": 0.166, "change": {"new_value": 2200.0, "old_value": 2121.5, "parameter": "Pool"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_sensitivity_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Chained multiplicative decomposition", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.166", "answer_float": 0.166, "change": {"new_value": 2200.0, "old_value": 2121.5, "parameter": "Pool"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Chained multiplicative decomposition", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.166", "answer_float": 0.166, "change": {"new_value": 2200.0, "old_value": 2121.5, "parameter": "Pool"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Chained multiplicative decomposition", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.166", "answer_float": 0.166, "change": {"new_value": 2200.0, "old_value": 2121.5, "parameter": "Pool"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Chained multiplicative decomposition", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF009_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.166", "answer_float": 0.166, "change": {"new_value": 2200.0, "old_value": 2121.5, "parameter": "Pool"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Chained multiplicative decomposition", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Pool=2121.5 g C/m2, Coef_K_Decom=0.001 d-1, F_temp=0.838 fraction, F_water=0.432 fraction, F_texture=0.77 fraction, F_pH=0.409 fraction, F_notill=0.9 fraction, Coef_K_Trans=0.264 fraction in carbon-nitrogen cycling soil carbon, what would the carbon-nitrogen cycling carbon pool respiration loss be if Pool were adjusted to 2200.0? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "27.35", "answer_float": 27.35, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "27.35", "answer_float": 27.35, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "27.35", "answer_float": 27.35, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "27.35", "answer_float": 27.35, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "27.35", "answer_float": 27.35, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "27.35", "answer_float": 27.35, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tair=30.2 C, calculate the Soil temperature from air temperature for a methane. The result should be in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "C"} +{"answer": "12.55", "answer_float": 12.55, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "12.55", "answer_float": 12.55, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "12.55", "answer_float": 12.55, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "12.55", "answer_float": 12.55, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "12.55", "answer_float": 12.55, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "12.55", "answer_float": 12.55, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "In phenology, the Critical daylength is obtained by chaining intermediate calculations. With inputs Po=10.5 h, PS=0.237 dimensionless, determine the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "h"} +{"answer": "275.26", "answer_float": 275.26, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "275.26", "answer_float": 275.26, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "275.26", "answer_float": 275.26, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "275.26", "answer_float": 275.26, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "275.26", "answer_float": 275.26, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "275.26", "answer_float": 275.26, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Subtraction and conversion", "mode": "direct", "precision": 2, "problem": "Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given DTGA=407.9 kg/ha/day, RM=96.1 kg/ha/day, RG=50.3 kg/ha/day, b=0.05 fraction, calculate the Daily dry matter increment for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "1.518", "answer_float": 1.518, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "1.518", "answer_float": 1.518, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子).\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "1.518", "answer_float": 1.518, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "1.518", "answer_float": 1.518, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "1.518", "answer_float": 1.518, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "1.518", "answer_float": 1.518, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF013", "formula_name_en": "carbon-nitrogen cycling crop temperature activity factor", "id": "CMF013_branch_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Beta temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling crop temperature activity factor in carbon-nitrogen cycling stress with these inputs: Td=33.8 C, Tmin_active=3.0 C, Topt_active=25.0 C, Tmax_active=42.0 C, Coef_T_M=0.76 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF013_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling stress", "template_id": 2, "unit": "response factor"} +{"answer": "9.179", "answer_float": 9.179, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "9.179", "answer_float": 9.179, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "9.179", "answer_float": 9.179, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "9.179", "answer_float": 9.179, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "9.179", "answer_float": 9.179, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "9.179", "answer_float": 9.179, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Ebl=4.01 g CH4/m2/day, P=15.119 g CH4/m2/day, CH4RiceEfC=0.386 fraction, W=4936.9 kg/ha, Wmax=12837.1 kg/ha, compute Total methane emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "0.629", "answer_float": 0.629, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.629", "answer_float": 0.629, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.629", "answer_float": 0.629, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.629", "answer_float": 0.629, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.629", "answer_float": 0.629, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.629", "answer_float": 0.629, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling respiration, the carbon-nitrogen cycling plant maintenance respiration is computed as part of carbon-nitrogen cycling respiration modeling. With Biom_plant_yesterday=221.0 g C/m2, Coef_Rm=0.0016 g C/g C/s, Tmn=25.3 C, DVI=1.15 dimensionless, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "12.27", "answer_float": 12.27, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t0_0001__C", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.27", "answer_float": 12.27, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t0_0001__K_name", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.27", "answer_float": 12.27, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t0_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.27", "answer_float": 12.27, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t0_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.27", "answer_float": 12.27, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t0_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "12.27", "answer_float": 12.27, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_chain_2_t0_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Power response", "mode": "chain_2", "precision": 2, "problem": "A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology module estimates Critical daylength using intermediate drivers. Use the provided inputs in sequence. The measurements are Po=10.5 h, PS=0.318 dimensionless. Compute the critical daylength and give the final numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "h"} +{"answer": "0.825", "answer_float": 0.825, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_branch_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.825", "answer_float": 0.825, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_branch_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.825", "answer_float": 0.825, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_branch_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.825", "answer_float": 0.825, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_branch_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.825", "answer_float": 0.825, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_branch_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.825", "answer_float": 0.825, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_branch_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise exponential", "mode": "branch", "precision": 3, "problem": "A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Eh=-133.0 mV. Determine the redox potential factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise linear stress", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response heat stress response. The measured inputs are Tmax=33.9 C, MaxT=35.1 C, ExtremeT=42.5 C. What is the simple heat stress response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "77.039", "answer_float": 77.039, "change": {"new_value": 0.9, "old_value": 0.78, "parameter": "SI"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "77.039", "answer_float": 77.039, "change": {"new_value": 0.9, "old_value": 0.78, "parameter": "SI"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root-exudate carbon substrate (根系分泌碳底物).\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "77.039", "answer_float": 77.039, "change": {"new_value": 0.9, "old_value": 0.78, "parameter": "SI"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "77.039", "answer_float": 77.039, "change": {"new_value": 0.9, "old_value": 0.78, "parameter": "SI"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "77.039", "answer_float": 77.039, "change": {"new_value": 0.9, "old_value": 0.78, "parameter": "SI"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "77.039", "answer_float": 77.039, "change": {"new_value": 0.9, "old_value": 0.78, "parameter": "SI"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF015", "formula_name_en": "Root-exudate carbon substrate", "id": "CMF015_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Power product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements VI=1.0 fraction, SI=0.78 fraction, W=5517.7 kg/ha in methane, what would the Root-exudate carbon substrate be if SI were adjusted to 0.9? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF015_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 32.7, "old_value": 29.7, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Multi-step sigmoid", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 32.7, "old_value": 29.7, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Multi-step sigmoid", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 32.7, "old_value": 29.7, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Multi-step sigmoid", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 32.7, "old_value": 29.7, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Multi-step sigmoid", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 32.7, "old_value": 29.7, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Multi-step sigmoid", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.09", "answer_float": 0.09, "change": {"new_value": 32.7, "old_value": 29.7, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Multi-step sigmoid", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=6390.0 kg/ha, B_above=12138.0 kg/ha, PDT=29.7 physiological days in dry matter partitioning, what would the Panicle partitioning index be if PDT were adjusted to 32.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.06", "answer_float": 0.06, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise logistic", "mode": "direct", "precision": 2, "problem": "Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.06", "answer_float": 0.06, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise logistic", "mode": "direct", "precision": 2, "problem": "Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: High-temperature fertility factor (高温结实因子).\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.06", "answer_float": 0.06, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise logistic", "mode": "direct", "precision": 2, "problem": "Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.06", "answer_float": 0.06, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise logistic", "mode": "direct", "precision": 2, "problem": "Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.06", "answer_float": 0.06, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise logistic", "mode": "direct", "precision": 2, "problem": "Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.06", "answer_float": 0.06, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF035", "formula_name_en": "High-temperature fertility factor", "id": "CMF035_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise logistic", "mode": "direct", "precision": 2, "problem": "Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=38.5 physiological days, Tmax_day=39.8 C, calculate the High-temperature fertility factor for a temperature stress. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF035_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 1, "unit": "fraction"} +{"answer": "0.92", "answer_float": 0.92, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "0.92", "answer_float": 0.92, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "0.92", "answer_float": 0.92, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "0.92", "answer_float": 0.92, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "0.92", "answer_float": 0.92, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "0.92", "answer_float": 0.92, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00013 ha/kg, AWL=7061.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "18.864", "answer_float": 18.864, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "18.864", "answer_float": 18.864, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "18.864", "answer_float": 18.864, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "18.864", "answer_float": 18.864, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "18.864", "answer_float": 18.864, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "18.864", "answer_float": 18.864, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.16 umol CO2/gN/s, Coef_beta=129.6 umol/m2/s, PAR=785.1 umol/m2/s, Leaf_N=1.36 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "5.811", "answer_float": 5.811, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "multiplier"} +{"answer": "5.811", "answer_float": 5.811, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "multiplier"} +{"answer": "5.811", "answer_float": 5.811, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "multiplier"} +{"answer": "5.811", "answer_float": 5.811, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "multiplier"} +{"answer": "5.811", "answer_float": 5.811, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "multiplier"} +{"answer": "5.811", "answer_float": 5.811, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate carbon-nitrogen cycling soil carbon temperature factor in carbon-nitrogen cycling soil carbon with these inputs: Temp=30.1 C, is_soil_temp=1.0 0/1. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "multiplier"} +{"answer": "0.290", "answer_float": 0.29, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.290", "answer_float": 0.29, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.290", "answer_float": 0.29, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.290", "answer_float": 0.29, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.290", "answer_float": 0.29, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.290", "answer_float": 0.29, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=51.3 physiological days, DTE=0.337 fraction, IE=0.73 dimensionless, RPE=0.26 fraction, BFF=0.86 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.758", "answer_float": 0.758, "change": {"new_value": 0.579, "old_value": 0.643, "parameter": "WaterC"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.758", "answer_float": 0.758, "change": {"new_value": 0.579, "old_value": 0.643, "parameter": "WaterC"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.758", "answer_float": 0.758, "change": {"new_value": 0.579, "old_value": 0.643, "parameter": "WaterC"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.758", "answer_float": 0.758, "change": {"new_value": 0.579, "old_value": 0.643, "parameter": "WaterC"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.758", "answer_float": 0.758, "change": {"new_value": 0.579, "old_value": 0.643, "parameter": "WaterC"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.758", "answer_float": 0.758, "change": {"new_value": 0.579, "old_value": 0.643, "parameter": "WaterC"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WaterC=0.643 fraction in methane, what would the Water-content decomposition factor be if WaterC were adjusted to 0.579? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.502", "answer_float": 0.502, "change": {"new_value": 32.2, "old_value": 29.3, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.502", "answer_float": 0.502, "change": {"new_value": 32.2, "old_value": 29.3, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.502", "answer_float": 0.502, "change": {"new_value": 32.2, "old_value": 29.3, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.502", "answer_float": 0.502, "change": {"new_value": 32.2, "old_value": 29.3, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.502", "answer_float": 0.502, "change": {"new_value": 32.2, "old_value": 29.3, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.502", "answer_float": 0.502, "change": {"new_value": 32.2, "old_value": 29.3, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Canopy extinction coefficient responds to parameter changes. Baseline: PDT=29.3 physiological days. After changing PDT from 29.3 to 32.2, compute the new Canopy extinction coefficient. Give the numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "dimensionless"} +{"answer": "0.654", "answer_float": 0.654, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.654", "answer_float": 0.654, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.654", "answer_float": 0.654, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.654", "answer_float": 0.654, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.654", "answer_float": 0.654, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "0.654", "answer_float": 0.654, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Daily physiological effectiveness in phenology with these inputs: PDT=46.1 physiological days, DTE=0.735 fraction, IE=0.8 dimensionless, RPE=0.7 fraction, BFF=0.89 dimensionless. The computation may follow different paths based on piecewise thresholds. Give the numeric result in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 2, "unit": "physiological days/day"} +{"answer": "19.87", "answer_float": 19.87, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "19.87", "answer_float": 19.87, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "19.87", "answer_float": 19.87, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "19.87", "answer_float": 19.87, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "19.87", "answer_float": 19.87, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "19.87", "answer_float": 19.87, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate Specific leaf area in leaf area growth with these inputs: GDD=774.4 C day, SLAc=146.5 cm2/g. The computation may follow different paths based on piecewise thresholds. Give the numeric result in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "cm2/g"} +{"answer": "0.976", "answer_float": 0.976, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.976", "answer_float": 0.976, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.976", "answer_float": 0.976, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.976", "answer_float": 0.976, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.976", "answer_float": 0.976, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.976", "answer_float": 0.976, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Water-content decomposition factor is computed as part of methane modeling. With WaterC=0.321 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "75.74", "answer_float": 75.74, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "75.74", "answer_float": 75.74, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "75.74", "answer_float": 75.74, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "75.74", "answer_float": 75.74, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "75.74", "answer_float": 75.74, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "75.74", "answer_float": 75.74, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response daily biomass increment is computed as part of environmental-response formula calculation. With RUE=1.99 g/MJ, fSolar=0.39 fraction, SRAD=24.3 MJ/m2/day, F_CO2=1.19 fraction, F_Temp=0.75 fraction, F_Water=0.8 fraction, F_Heat=0.45 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "14.89", "answer_float": 14.89, "change": {"new_value": 13.8, "old_value": 12.0, "parameter": "Tair"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "14.89", "answer_float": 14.89, "change": {"new_value": 13.8, "old_value": 12.0, "parameter": "Tair"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "14.89", "answer_float": 14.89, "change": {"new_value": 13.8, "old_value": 12.0, "parameter": "Tair"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "14.89", "answer_float": 14.89, "change": {"new_value": 13.8, "old_value": 12.0, "parameter": "Tair"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "14.89", "answer_float": 14.89, "change": {"new_value": 13.8, "old_value": 12.0, "parameter": "Tair"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "14.89", "answer_float": 14.89, "change": {"new_value": 13.8, "old_value": 12.0, "parameter": "Tair"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 2, "problem": "In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil temperature from air temperature responds to parameter changes. Baseline: Tair=12.0 C. After changing Tair from 12.0 to 13.8, compute the new Soil temperature from air temperature. Give the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "254.39", "answer_float": 254.39, "change": {"new_value": 5.23, "old_value": 5.81, "parameter": "LAI"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "umol/m2/s"} +{"answer": "254.39", "answer_float": 254.39, "change": {"new_value": 5.23, "old_value": 5.81, "parameter": "LAI"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "umol/m2/s"} +{"answer": "254.39", "answer_float": 254.39, "change": {"new_value": 5.23, "old_value": 5.81, "parameter": "LAI"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "umol/m2/s"} +{"answer": "254.39", "answer_float": 254.39, "change": {"new_value": 5.23, "old_value": 5.81, "parameter": "LAI"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "umol/m2/s"} +{"answer": "254.39", "answer_float": 254.39, "change": {"new_value": 5.23, "old_value": 5.81, "parameter": "LAI"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "umol/m2/s"} +{"answer": "254.39", "answer_float": 254.39, "change": {"new_value": 5.23, "old_value": 5.81, "parameter": "LAI"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Radi=583.4 W/m2, LAI=5.81 m2/m2, Coef_rou=0.18 fraction, Coef_K=0.71 dimensionless in carbon-nitrogen cycling photosynthesis, what would the carbon-nitrogen cycling canopy absorbed PAR be if LAI were adjusted to 5.23? Provide the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "umol/m2/s"} +{"answer": "5.576", "answer_float": 5.576, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t0_0001__C", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "5.576", "answer_float": 5.576, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t0_0001__K_name", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "5.576", "answer_float": 5.576, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t0_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "5.576", "answer_float": 5.576, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t0_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "5.576", "answer_float": 5.576, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t0_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "5.576", "answer_float": 5.576, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t0_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling respiration involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Biom_plant_yesterday=641.5 g C/m2, Coef_Rm=0.0033 g C/g C/s, Tmn=30.2 C, DVI=1.12 dimensionless. Determine the carbon-nitrogen cycling plant maintenance respiration and report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.433", "answer_float": 0.433, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t0_0000__C", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.433", "answer_float": 0.433, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t0_0000__K_name", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.433", "answer_float": 0.433, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t0_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.433", "answer_float": 0.433, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t0_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.433", "answer_float": 0.433, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t0_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.433", "answer_float": 0.433, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t0_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Methane production using intermediate drivers. Use the provided inputs in sequence. The measurements are FEh=0.128 fraction, TI=0.774 fraction, Cr=10.96 g C/m2/day, Com=4.036 g C/m2/day. Compute the methane production and give the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g CH4/m2/day"} +{"answer": "0.90", "answer_float": 0.9, "change": {"new_value": 42.4, "old_value": 47.1, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.90", "answer_float": 0.9, "change": {"new_value": 42.4, "old_value": 47.1, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Shoot partitioning index (地上部分配指数).\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.90", "answer_float": 0.9, "change": {"new_value": 42.4, "old_value": 47.1, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.90", "answer_float": 0.9, "change": {"new_value": 42.4, "old_value": 47.1, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.90", "answer_float": 0.9, "change": {"new_value": 42.4, "old_value": 47.1, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "0.90", "answer_float": 0.9, "change": {"new_value": 42.4, "old_value": 47.1, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A dry matter partitioning calculation involves sensitivity analysis. The current measurements are PDT=47.1 physiological days. A field technician asks what the shoot partitioning index would become if PDT changed to 42.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 0, "unit": "fraction"} +{"answer": "4.656", "answer_float": 4.656, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "4.656", "answer_float": 4.656, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "4.656", "answer_float": 4.656, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "4.656", "answer_float": 4.656, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "4.656", "answer_float": 4.656, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "4.656", "answer_float": 4.656, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Q10 with DVI correction", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling respiration, compute the carbon-nitrogen cycling plant maintenance respiration. The inputs are Biom_plant_yesterday=757.9 g C/m2, Coef_Rm=0.003 g C/g C/s, Tmn=23.3 C, DVI=0.76 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "2.90", "answer_float": 2.9, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "2.90", "answer_float": 2.9, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "2.90", "answer_float": 2.9, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "2.90", "answer_float": 2.9, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "2.90", "answer_float": 2.9, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "2.90", "answer_float": 2.9, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Tmean=11.1 C, Tbase=8.2 C, calculate the environmental-response daily thermal time for an environmental-response formula system. The result should be in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "3.15", "answer_float": 3.15, "change": {"new_value": 8500.0, "old_value": 8345.0, "parameter": "AWL"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "m2/m2"} +{"answer": "3.15", "answer_float": 3.15, "change": {"new_value": 8500.0, "old_value": 8345.0, "parameter": "AWL"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "m2/m2"} +{"answer": "3.15", "answer_float": 3.15, "change": {"new_value": 8500.0, "old_value": 8345.0, "parameter": "AWL"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "m2/m2"} +{"answer": "3.15", "answer_float": 3.15, "change": {"new_value": 8500.0, "old_value": 8345.0, "parameter": "AWL"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "m2/m2"} +{"answer": "3.15", "answer_float": 3.15, "change": {"new_value": 8500.0, "old_value": 8345.0, "parameter": "AWL"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "m2/m2"} +{"answer": "3.15", "answer_float": 3.15, "change": {"new_value": 8500.0, "old_value": 8345.0, "parameter": "AWL"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In leaf area growth, we analyze how Leaf area index from SLA responds to parameter changes. Baseline: SLA=0.00037 ha/kg, AWL=8345.0 kg/ha. After changing AWL from 8345.0 to 8500.0, compute the new Leaf area index from SLA. Give the numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "m2/m2"} +{"answer": "0.0065", "answer_float": 0.0065, "change": {"new_value": 25.9, "old_value": 28.8, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Linear decline", "mode": "sensitivity", "precision": 4, "problem": "A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "d-1"} +{"answer": "0.0065", "answer_float": 0.0065, "change": {"new_value": 25.9, "old_value": 28.8, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Linear decline", "mode": "sensitivity", "precision": 4, "problem": "A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "d-1"} +{"answer": "0.0065", "answer_float": 0.0065, "change": {"new_value": 25.9, "old_value": 28.8, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Linear decline", "mode": "sensitivity", "precision": 4, "problem": "A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "d-1"} +{"answer": "0.0065", "answer_float": 0.0065, "change": {"new_value": 25.9, "old_value": 28.8, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Linear decline", "mode": "sensitivity", "precision": 4, "problem": "A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "d-1"} +{"answer": "0.0065", "answer_float": 0.0065, "change": {"new_value": 25.9, "old_value": 28.8, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Linear decline", "mode": "sensitivity", "precision": 4, "problem": "A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "d-1"} +{"answer": "0.0065", "answer_float": 0.0065, "change": {"new_value": 25.9, "old_value": 28.8, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Linear decline", "mode": "sensitivity", "precision": 4, "problem": "A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A respiration calculation involves sensitivity analysis. The current measurements are PDT=28.8 physiological days. A field technician asks what the maintenance respiration coefficient at optimum temperature would become if PDT changed to 25.9. Report the new final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 0, "unit": "d-1"} +{"answer": "0.652", "answer_float": 0.652, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_chain_2_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.652", "answer_float": 0.652, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_chain_2_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.652", "answer_float": 0.652, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_chain_2_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.652", "answer_float": 0.652, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_chain_2_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.652", "answer_float": 0.652, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_chain_2_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.652", "answer_float": 0.652, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_chain_2_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Fresh organic matter carbon decomposition is obtained by chaining intermediate calculations. With inputs WI=0.279 fraction, SI=0.65 fraction, TI=0.68 fraction, OMN=195.9 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.20", "answer_float": 0.2, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_branch_t1_0001__C", "is_gold": false, "level": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "precision": 2, "problem": "For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.20", "answer_float": 0.2, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_branch_t1_0001__K_name", "is_gold": false, "level": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "precision": 2, "problem": "For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Panicle partitioning index (穗分配指数).\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.20", "answer_float": 0.2, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_branch_t1_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "precision": 2, "problem": "For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.20", "answer_float": 0.2, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_branch_t1_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "precision": 2, "problem": "For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.20", "answer_float": 0.2, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_branch_t1_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "precision": 2, "problem": "For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.20", "answer_float": 0.2, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF042", "formula_name_en": "Panicle partitioning index", "id": "CMF042_branch_t1_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Multi-step sigmoid", "mode": "branch", "precision": 2, "problem": "For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For dry matter partitioning, compute the Panicle partitioning index. The inputs are Y_grain=4369.0 kg/ha, B_above=13156.0 kg/ha, PDT=39.3 physiological days. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF042_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "40.80", "answer_float": 40.8, "change": {"new_value": 5430.0, "old_value": 4936.0, "parameter": "GrainYield"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "40.80", "answer_float": 40.8, "change": {"new_value": 5430.0, "old_value": 4936.0, "parameter": "GrainYield"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "40.80", "answer_float": 40.8, "change": {"new_value": 5430.0, "old_value": 4936.0, "parameter": "GrainYield"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "40.80", "answer_float": 40.8, "change": {"new_value": 5430.0, "old_value": 4936.0, "parameter": "GrainYield"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "40.80", "answer_float": 40.8, "change": {"new_value": 5430.0, "old_value": 4936.0, "parameter": "GrainYield"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "40.80", "answer_float": 40.8, "change": {"new_value": 5430.0, "old_value": 4936.0, "parameter": "GrainYield"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements GrainYield=4936.0 kg/ha, NUptake=133.1 kg N/ha in nitrogen, what would the Nitrogen use efficiency be if GrainYield were adjusted to 5430.0? Provide the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 1, "unit": "kg grain/kg N"} +{"answer": "68.1", "answer_float": 68.1, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t0_0000__C", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "C day"} +{"answer": "68.1", "answer_float": 68.1, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t0_0000__K_name", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Emergence thermal requirement (出苗热量需求).\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "C day"} +{"answer": "68.1", "answer_float": 68.1, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t0_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "C day"} +{"answer": "68.1", "answer_float": 68.1, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t0_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "C day"} +{"answer": "68.1", "answer_float": 68.1, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t0_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "C day"} +{"answer": "68.1", "answer_float": 68.1, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF028", "formula_name_en": "Emergence thermal requirement", "id": "CMF028_direct_t0_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 1, "problem": "A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Emergence thermal requirement. The measured inputs are SDEPTH=3.3 cm. What is the emergence thermal requirement? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF028_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "C day"} +{"answer": "0.000", "answer_float": 0.0, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_branch_t0_0001__C", "is_gold": false, "level": 4, "math_type": "Piecewise linear", "mode": "branch", "precision": 3, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_branch_t0_0001__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise linear", "mode": "branch", "precision": 3, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_branch_t0_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise linear", "mode": "branch", "precision": 3, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_branch_t0_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise linear", "mode": "branch", "precision": 3, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_branch_t0_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise linear", "mode": "branch", "precision": 3, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_branch_t0_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise linear", "mode": "branch", "precision": 3, "problem": "A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Tmean=3.3 C, Tbase=8.4 C, Topt=22.7 C. Determine the simple temperature response and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 42.0, "old_value": 41.5, "parameter": "t_soil"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 42.0, "old_value": 41.5, "parameter": "t_soil"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane temperature index (甲烷温度指数).\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 42.0, "old_value": 41.5, "parameter": "t_soil"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 42.0, "old_value": 41.5, "parameter": "t_soil"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 42.0, "old_value": 41.5, "parameter": "t_soil"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": {"new_value": 42.0, "old_value": 41.5, "parameter": "t_soil"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF023", "formula_name_en": "Methane temperature index", "id": "CMF023_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Q10 with temperature clipping", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Methane temperature index responds to parameter changes. Baseline: Q10=3.41 dimensionless, t_soil=41.5 C. After changing t_soil from 41.5 to 42.0, compute the new Methane temperature index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF023_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "multiplier"} +{"answer": "0.66", "answer_float": 0.66, "change": {"new_value": 8095.0, "old_value": 7039.0, "parameter": "Y_grain"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.66", "answer_float": 0.66, "change": {"new_value": 8095.0, "old_value": 7039.0, "parameter": "Y_grain"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Harvest index (收获指数).\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.66", "answer_float": 0.66, "change": {"new_value": 8095.0, "old_value": 7039.0, "parameter": "Y_grain"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.66", "answer_float": 0.66, "change": {"new_value": 8095.0, "old_value": 7039.0, "parameter": "Y_grain"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.66", "answer_float": 0.66, "change": {"new_value": 8095.0, "old_value": 7039.0, "parameter": "Y_grain"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.66", "answer_float": 0.66, "change": {"new_value": 8095.0, "old_value": 7039.0, "parameter": "Y_grain"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Y_grain=7039.0 kg/ha, B_above=12285.0 kg/ha in dry matter partitioning, what would the Harvest index be if Y_grain were adjusted to 8095.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_branch_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise linear stress", "mode": "branch", "precision": 3, "problem": "Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_branch_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise linear stress", "mode": "branch", "precision": 3, "problem": "Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_branch_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise linear stress", "mode": "branch", "precision": 3, "problem": "Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_branch_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise linear stress", "mode": "branch", "precision": 3, "problem": "Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_branch_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise linear stress", "mode": "branch", "precision": 3, "problem": "Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_branch_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise linear stress", "mode": "branch", "precision": 3, "problem": "Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Calculate environmental-response heat stress response in environmental-response formula system with these inputs: Tmax=32.7 C, MaxT=34.5 C, ExtremeT=43.3 C. The computation may follow different paths based on piecewise thresholds. Give the numeric result in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_branch_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.648", "answer_float": 0.648, "change": {"new_value": 22.2, "old_value": 20.2, "parameter": "Tmean"}, "condition": "C", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.648", "answer_float": 0.648, "change": {"new_value": 22.2, "old_value": 20.2, "parameter": "Tmean"}, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.648", "answer_float": 0.648, "change": {"new_value": 22.2, "old_value": 20.2, "parameter": "Tmean"}, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.648", "answer_float": 0.648, "change": {"new_value": 22.2, "old_value": 20.2, "parameter": "Tmean"}, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.648", "answer_float": 0.648, "change": {"new_value": 22.2, "old_value": 20.2, "parameter": "Tmean"}, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.648", "answer_float": 0.648, "change": {"new_value": 22.2, "old_value": 20.2, "parameter": "Tmean"}, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=20.2 C, Tbase=9.5 C, Topt=29.1 C in environmental-response formula system, what would the environmental-response temperature response be if Tmean were adjusted to 22.2? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.14", "answer_float": 0.14, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_chain_2_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Residual from shoot partitioning", "mode": "chain_2", "precision": 2, "problem": "In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.14", "answer_float": 0.14, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_chain_2_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Residual from shoot partitioning", "mode": "chain_2", "precision": 2, "problem": "In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.14", "answer_float": 0.14, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_chain_2_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Residual from shoot partitioning", "mode": "chain_2", "precision": 2, "problem": "In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.14", "answer_float": 0.14, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_chain_2_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Residual from shoot partitioning", "mode": "chain_2", "precision": 2, "problem": "In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.14", "answer_float": 0.14, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Residual from shoot partitioning", "mode": "chain_2", "precision": 2, "problem": "In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.14", "answer_float": 0.14, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Residual from shoot partitioning", "mode": "chain_2", "precision": 2, "problem": "In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, the Root partitioning index is obtained by chaining intermediate calculations. With inputs PDT=30.5 physiological days, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.157", "answer_float": 0.157, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.157", "answer_float": 0.157, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.157", "answer_float": 0.157, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.157", "answer_float": 0.157, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.157", "answer_float": 0.157, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "0.157", "answer_float": 0.157, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling water calculation estimates carbon-nitrogen cycling crop water stress factor. The measured inputs are Soil_Water=0.102 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. What is the carbon-nitrogen cycling crop water stress factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 0, "unit": "fraction"} +{"answer": "38.61", "answer_float": 38.61, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "38.61", "answer_float": 38.61, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "38.61", "answer_float": 38.61, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "38.61", "answer_float": 38.61, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "38.61", "answer_float": 38.61, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "38.61", "answer_float": 38.61, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Rg=0.26 fraction, DTGA=148.5 kg/ha/day, calculate the Growth respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "0.736", "answer_float": 0.736, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_direct_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Double logistic", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.736", "answer_float": 0.736, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_direct_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Double logistic", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response solar interception fraction (环境响应光截获比例).\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.736", "answer_float": 0.736, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_direct_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Double logistic", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.736", "answer_float": 0.736, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_direct_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Double logistic", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.736", "answer_float": 0.736, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_direct_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Double logistic", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.736", "answer_float": 0.736, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF059", "formula_name_en": "environmental-response solar interception fraction", "id": "CMF059_direct_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Double logistic", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response solar interception fraction is computed as part of environmental-response formula calculation. With fSolarMax=0.89 fraction, TT=1058.6 C day, I50A=388.4 C day, I50B=398.6 C day, Tsum=1779.5 C day, fSolar_water=0.86 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF059_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "12.42", "answer_float": 12.42, "change": {"new_value": 0.27, "old_value": 0.235, "parameter": "PS"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "h"} +{"answer": "12.42", "answer_float": 12.42, "change": {"new_value": 0.27, "old_value": 0.235, "parameter": "PS"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Critical daylength (临界日长).\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "h"} +{"answer": "12.42", "answer_float": 12.42, "change": {"new_value": 0.27, "old_value": 0.235, "parameter": "PS"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "h"} +{"answer": "12.42", "answer_float": 12.42, "change": {"new_value": 0.27, "old_value": 0.235, "parameter": "PS"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "h"} +{"answer": "12.42", "answer_float": 12.42, "change": {"new_value": 0.27, "old_value": 0.235, "parameter": "PS"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "h"} +{"answer": "12.42", "answer_float": 12.42, "change": {"new_value": 0.27, "old_value": 0.235, "parameter": "PS"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF040", "formula_name_en": "Critical daylength", "id": "CMF040_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Power response", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Po=10.5 h, PS=0.235 dimensionless in phenology, what would the Critical daylength be if PS were adjusted to 0.27? Provide the numeric value in h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF040_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "h"} +{"answer": "101.31", "answer_float": 101.31, "change": {"new_value": 0.23, "old_value": 0.2, "parameter": "Rg"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "101.31", "answer_float": 101.31, "change": {"new_value": 0.23, "old_value": 0.2, "parameter": "Rg"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Growth respiration (生长呼吸).\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "101.31", "answer_float": 101.31, "change": {"new_value": 0.23, "old_value": 0.2, "parameter": "Rg"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "101.31", "answer_float": 101.31, "change": {"new_value": 0.23, "old_value": 0.2, "parameter": "Rg"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "101.31", "answer_float": 101.31, "change": {"new_value": 0.23, "old_value": 0.2, "parameter": "Rg"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "101.31", "answer_float": 101.31, "change": {"new_value": 0.23, "old_value": 0.2, "parameter": "Rg"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF046", "formula_name_en": "Growth respiration", "id": "CMF046_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Growth respiration responds to parameter changes. Baseline: Rg=0.2 fraction, DTGA=440.5 kg/ha/day. After changing Rg from 0.2 to 0.23, compute the new Growth respiration. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF046_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "0.024", "answer_float": 0.024, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_chain_2_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "precision": 3, "problem": "Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "0.024", "answer_float": 0.024, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_chain_2_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "precision": 3, "problem": "Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane ebullition emission (甲烷冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "0.024", "answer_float": 0.024, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_chain_2_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "precision": 3, "problem": "Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "0.024", "answer_float": 0.024, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_chain_2_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "precision": 3, "problem": "Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "0.024", "answer_float": 0.024, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_chain_2_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "precision": 3, "problem": "Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "0.024", "answer_float": 0.024, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_chain_2_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Logarithmic minimum", "mode": "chain_2", "precision": 3, "problem": "Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given P=16.848 g CH4/m2/day, t_soil=33.0 C, Wr=1725.2 kg/ha, compute Methane ebullition emission in methane. This requires sequential calculation through intermediate variables. Report the final numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "448.60", "answer_float": 448.6, "change": {"new_value": 2.74, "old_value": 2.49, "parameter": "LAI"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "448.60", "answer_float": 448.6, "change": {"new_value": 2.74, "old_value": 2.49, "parameter": "LAI"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "448.60", "answer_float": 448.6, "change": {"new_value": 2.74, "old_value": 2.49, "parameter": "LAI"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "448.60", "answer_float": 448.6, "change": {"new_value": 2.74, "old_value": 2.49, "parameter": "LAI"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "448.60", "answer_float": 448.6, "change": {"new_value": 2.74, "old_value": 2.49, "parameter": "LAI"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "448.60", "answer_float": 448.6, "change": {"new_value": 2.74, "old_value": 2.49, "parameter": "LAI"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Exponential ratio", "mode": "sensitivity", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation involves sensitivity analysis. The current measurements are Radi=588.1 W/m2, LAI=2.49 m2/m2, Coef_rou=0.12 fraction, Coef_K=0.74 dimensionless. A field technician asks what the carbon-nitrogen cycling canopy absorbed par would become if LAI changed to 2.74. Report the new final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "0.65", "answer_float": 0.65, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.65", "answer_float": 0.65, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Stem partitioning index (茎鞘分配指数).\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.65", "answer_float": 0.65, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.65", "answer_float": 0.65, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.65", "answer_float": 0.65, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.65", "answer_float": 0.65, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF045", "formula_name_en": "Stem partitioning index", "id": "CMF045_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Residual with piecewise components", "mode": "direct", "precision": 2, "problem": "Given PDT=35.1 physiological days, Y_grain=5337.0 kg/ha, B_above=16042.0 kg/ha, calculate the Stem partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF045_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.015", "answer_float": 0.015, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Logarithmic minimum", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.015", "answer_float": 0.015, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Logarithmic minimum", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include 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��冒泡排放).\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.015", "answer_float": 0.015, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Logarithmic minimum", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.015", "answer_float": 0.015, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Logarithmic minimum", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.015", "answer_float": 0.015, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Logarithmic minimum", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.015", "answer_float": 0.015, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF016", "formula_name_en": "Methane ebullition emission", "id": "CMF016_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Logarithmic minimum", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane ebullition emission is computed as part of methane modeling. With P=8.889 g CH4/m2/day, t_soil=34.8 C, Wr=1500.3 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF016_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.134", "answer_float": 6.134, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.134", "answer_float": 6.134, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.134", "answer_float": 6.134, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.134", "answer_float": 6.134, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.134", "answer_float": 6.134, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "6.134", "answer_float": 6.134, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_chain_2_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Chained sum", "mode": "chain_2", "precision": 3, "problem": "In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Total methane emission is obtained by chaining intermediate calculations. With inputs Ebl=5.626 g CH4/m2/day, P=2.569 g CH4/m2/day, CH4RiceEfC=0.244 fraction, W=6214.2 kg/ha, Wmax=10951.2 kg/ha, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "35.555", "answer_float": 35.555, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "35.555", "answer_float": 35.555, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "35.555", "answer_float": 35.555, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "35.555", "answer_float": 35.555, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "35.555", "answer_float": 35.555, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "35.555", "answer_float": 35.555, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling optimal photosynthetic rate is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Coef_alfa=16.85 umol CO2/gN/s, Coef_beta=158.8 umol/m2/s, PAR=261.8 umol/m2/s, Leaf_N=3.39 gN/m2, determine the numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol CO2/m2/s"} +{"answer": "2.161", "answer_float": 2.161, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_direct_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Q10 temperature response", "mode": "direct", "precision": 3, "problem": "Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "multiplier"} +{"answer": "2.161", "answer_float": 2.161, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_direct_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Q10 temperature response", "mode": "direct", "precision": 3, "problem": "Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "multiplier"} +{"answer": "2.161", "answer_float": 2.161, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_direct_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Q10 temperature response", "mode": "direct", "precision": 3, "problem": "Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "multiplier"} +{"answer": "2.161", "answer_float": 2.161, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_direct_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Q10 temperature response", "mode": "direct", "precision": 3, "problem": "Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "multiplier"} +{"answer": "2.161", "answer_float": 2.161, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_direct_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Q10 temperature response", "mode": "direct", "precision": 3, "problem": "Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "multiplier"} +{"answer": "2.161", "answer_float": 2.161, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_direct_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Q10 temperature response", "mode": "direct", "precision": 3, "problem": "Given Temp=18.8 C, is_soil_temp=1.0 0/1, calculate the carbon-nitrogen cycling soil carbon temperature factor for a carbon-nitrogen cycling soil carbon. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF011_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 1, "unit": "multiplier"} +{"answer": "0.698", "answer_float": 0.698, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0001__C", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "0.698", "answer_float": 0.698, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0001__K_name", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "0.698", "answer_float": 0.698, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "0.698", "answer_float": 0.698, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "0.698", "answer_float": 0.698, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "0.698", "answer_float": 0.698, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=5.9 C, is_soil_temp=1.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "0.814", "answer_float": 0.814, "change": {"new_value": 5.59, "old_value": 4.86, "parameter": "pH"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_sensitivity_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Logistic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.814", "answer_float": 0.814, "change": {"new_value": 5.59, "old_value": 4.86, "parameter": "pH"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_sensitivity_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Logistic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子).\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.814", "answer_float": 0.814, "change": {"new_value": 5.59, "old_value": 4.86, "parameter": "pH"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Logistic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.814", "answer_float": 0.814, "change": {"new_value": 5.59, "old_value": 4.86, "parameter": "pH"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Logistic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF010_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.814", "answer_float": 0.814, "change": {"new_value": 5.59, "old_value": 4.86, "parameter": "pH"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Logistic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF010_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "0.814", "answer_float": 0.814, "change": {"new_value": 5.59, "old_value": 4.86, "parameter": "pH"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF010", "formula_name_en": "carbon-nitrogen cycling soil pH decomposition factor", "id": "CMF010_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Logistic", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, we analyze how carbon-nitrogen cycling soil pH decomposition factor responds to parameter changes. Baseline: pH=4.86 pH. After changing pH from 4.86 to 5.59, compute the new carbon-nitrogen cycling soil pH decomposition factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF010_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise exponential decay", "mode": "direct", "precision": 2, "problem": "Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=15.7 physiological days, a=0.023 dimensionless, calculate the Physiological age factor for a photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "25.45", "answer_float": 25.45, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "25.45", "answer_float": 25.45, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "25.45", "answer_float": 25.45, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "25.45", "answer_float": 25.45, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "25.45", "answer_float": 25.45, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "25.45", "answer_float": 25.45, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil temperature from air temperature is computed as part of methane modeling. With Tair=27.7 C, determine the numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "C"} +{"answer": "0.374", "answer_float": 0.374, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.374", "answer_float": 0.374, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.374", "answer_float": 0.374, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.374", "answer_float": 0.374, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.374", "answer_float": 0.374, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.374", "answer_float": 0.374, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Daily physiological effectiveness. The measured inputs are PDT=2.4 physiological days, DTE=0.331 fraction, IE=1.13 dimensionless, RPE=0.64 fraction, BFF=0.69 dimensionless. What is the daily physiological effectiveness? Give the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "2.086", "answer_float": 2.086, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "2.086", "answer_float": 2.086, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "2.086", "answer_float": 2.086, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "2.086", "answer_float": 2.086, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "2.086", "answer_float": 2.086, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "2.086", "answer_float": 2.086, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is computed as part of carbon-nitrogen cycling photosynthesis modeling. With Photo_Rate_opt=23.71 umol CO2/m2/s, F_T=0.42 fraction, F_Water=0.36 fraction, F_CO2=1.21 fraction, Day_Length_hour=11.13 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.972", "answer_float": 0.972, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.972", "answer_float": 0.972, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.972", "answer_float": 0.972, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.972", "answer_float": 0.972, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.972", "answer_float": 0.972, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.972", "answer_float": 0.972, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.459 fraction, SI=1.33 fraction, TI=0.435 fraction, OMS=1220.1 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "135.57", "answer_float": 135.57, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "cm2/g"} +{"answer": "135.57", "answer_float": 135.57, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "cm2/g"} +{"answer": "135.57", "answer_float": 135.57, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "cm2/g"} +{"answer": "135.57", "answer_float": 135.57, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "cm2/g"} +{"answer": "135.57", "answer_float": 135.57, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "cm2/g"} +{"answer": "135.57", "answer_float": 135.57, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise polynomial", "mode": "branch", "precision": 2, "problem": "For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For leaf area growth, compute the Specific leaf area. The inputs are GDD=597.9 C day, SLAc=155.4 cm2/g. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "cm2/g"} +{"answer": "1.519", "answer_float": 1.519, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t0_0001__C", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.519", "answer_float": 1.519, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t0_0001__K_name", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.519", "answer_float": 1.519, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t0_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.519", "answer_float": 1.519, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t0_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.519", "answer_float": 1.519, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t0_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root-exudate carbon substrate (根系分泌碳底物)\nFormula (LaTeX): C_r=0.0018VI\\cdot SI\\cdot W^{1.25}\nFormula: Cr = 0.0018 * VI * SI * W ** 1.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- VI: variety index; unit=fraction; typical range=[0.7, 1.3]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- W: shoot biomass; unit=kg/ha; typical range=[500.0, 8500.0]\nPhysiological category: Methane\nMathematical form: Power product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "1.519", "answer_float": 1.519, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_chain_2_t0_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Multiplicative", "mode": "chain_2", "precision": 3, "problem": "A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane module estimates Soil organic matter carbon decomposition using intermediate drivers. Use the provided inputs in sequence. The measurements are WI=0.541 fraction, SI=0.92 fraction, TI=0.658 fraction, OMS=1545.9 g C/m2. Compute the soil organic matter carbon decomposition and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_chain_2_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "28.104", "answer_float": 28.104, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "28.104", "answer_float": 28.104, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "28.104", "answer_float": 28.104, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "28.104", "answer_float": 28.104, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "28.104", "answer_float": 28.104, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "28.104", "answer_float": 28.104, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=53.96 umol CO2/m2/s, F_T=0.95 fraction, F_Water=0.69 fraction, F_CO2=1.31 fraction, Day_Length_hour=14.04 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "3.42", "answer_float": 3.42, "change": {"new_value": 8153.0, "old_value": 7090.0, "parameter": "AWL"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "3.42", "answer_float": 3.42, "change": {"new_value": 8153.0, "old_value": 7090.0, "parameter": "AWL"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "3.42", "answer_float": 3.42, "change": {"new_value": 8153.0, "old_value": 7090.0, "parameter": "AWL"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "3.42", "answer_float": 3.42, "change": {"new_value": 8153.0, "old_value": 7090.0, "parameter": "AWL"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "3.42", "answer_float": 3.42, "change": {"new_value": 8153.0, "old_value": 7090.0, "parameter": "AWL"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "3.42", "answer_float": 3.42, "change": {"new_value": 8153.0, "old_value": 7090.0, "parameter": "AWL"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Product", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are SLA=0.00042 ha/kg, AWL=7090.0 kg/ha. A field technician asks what the leaf area index from sla would become if AWL changed to 8153.0. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "11.71", "answer_float": 11.71, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Product with exponential saturation", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "11.71", "answer_float": 11.71, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Product with exponential saturation", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Intercepted photosynthetically active radiation (截获光合有效辐射).\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "11.71", "answer_float": 11.71, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Product with exponential saturation", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "11.71", "answer_float": 11.71, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Product with exponential saturation", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "11.71", "answer_float": 11.71, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Product with exponential saturation", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "11.71", "answer_float": 11.71, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Product with exponential saturation", "mode": "direct", "precision": 2, "problem": "A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation estimates Intercepted photosynthetically active radiation. The measured inputs are PAR=14.18 MJ/m2/day, k=0.62 dimensionless, LAI=2.82 m2/m2. What is the intercepted photosynthetically active radiation? Give the final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "1.07", "answer_float": 1.07, "change": {"new_value": 0.64, "old_value": 0.56, "parameter": "beta"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.07", "answer_float": 1.07, "change": {"new_value": 0.64, "old_value": 0.56, "parameter": "beta"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.07", "answer_float": 1.07, "change": {"new_value": 0.64, "old_value": 0.56, "parameter": "beta"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.07", "answer_float": 1.07, "change": {"new_value": 0.64, "old_value": 0.56, "parameter": "beta"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.07", "answer_float": 1.07, "change": {"new_value": 0.64, "old_value": 0.56, "parameter": "beta"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.07", "answer_float": 1.07, "change": {"new_value": 0.64, "old_value": 0.56, "parameter": "beta"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=377.0 ppm, beta=0.56 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.64. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "26.59", "answer_float": 26.59, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "26.59", "answer_float": 26.59, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil temperature from air temperature (土壤温度估算).\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "26.59", "answer_float": 26.59, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "26.59", "answer_float": 26.59, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "26.59", "answer_float": 26.59, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "26.59", "answer_float": 26.59, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF024", "formula_name_en": "Soil temperature from air temperature", "id": "CMF024_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear", "mode": "direct", "precision": 2, "problem": "A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation estimates Soil temperature from air temperature. The measured inputs are Tair=29.2 C. What is the soil temperature from air temperature? Give the final numeric value in C.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF024_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "C"} +{"answer": "0.173", "answer_float": 0.173, "change": {"new_value": 0.41, "old_value": 0.45, "parameter": "RPE"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.173", "answer_float": 0.173, "change": {"new_value": 0.41, "old_value": 0.45, "parameter": "RPE"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.173", "answer_float": 0.173, "change": {"new_value": 0.41, "old_value": 0.45, "parameter": "RPE"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.173", "answer_float": 0.173, "change": {"new_value": 0.41, "old_value": 0.45, "parameter": "RPE"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.173", "answer_float": 0.173, "change": {"new_value": 0.41, "old_value": 0.45, "parameter": "RPE"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.173", "answer_float": 0.173, "change": {"new_value": 0.41, "old_value": 0.45, "parameter": "RPE"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Stage-dependent product", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=8.3 physiological days, DTE=0.422 fraction, IE=0.67 dimensionless, RPE=0.45 fraction, BFF=1.07 dimensionless in phenology, what would the Daily physiological effectiveness be if RPE were adjusted to 0.41? Provide the numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "1.454", "answer_float": 1.454, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.454", "answer_float": 1.454, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.454", "answer_float": 1.454, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.454", "answer_float": 1.454, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.454", "answer_float": 1.454, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.454", "answer_float": 1.454, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling CO2 fertilization factor. The measured inputs are CO2=613.0 ppm, Coef_B_CO2=0.77 dimensionless. What is the carbon-nitrogen cycling co2 fertilization factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "17.20", "answer_float": 17.2, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "kg CO2/ha/h"} +{"answer": "17.20", "answer_float": 17.2, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "kg CO2/ha/h"} +{"answer": "17.20", "answer_float": 17.2, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "kg CO2/ha/h"} +{"answer": "17.20", "answer_float": 17.2, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "kg CO2/ha/h"} +{"answer": "17.20", "answer_float": 17.2, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "kg CO2/ha/h"} +{"answer": "17.20", "answer_float": 17.2, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative with minimum stress", "mode": "chain_2", "precision": 2, "problem": "In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, the Actual maximum assimilation rate is obtained by chaining intermediate calculations. With inputs Am=28.0 kg CO2/ha/h, FCO2=1.16 fraction, FPA=0.85 fraction, FT=0.89 fraction, NNI=0.84 fraction, WDF=0.7 fraction, determine the numeric value in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "kg CO2/ha/h"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "branch", "precision": 2, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "branch", "precision": 2, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative photoperiod effectiveness (相对光周期效应).\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "branch", "precision": 2, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "branch", "precision": 2, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "branch", "precision": 2, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF050", "formula_name_en": "Relative photoperiod effectiveness", "id": "CMF050_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise quadratic", "mode": "branch", "precision": 2, "problem": "For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For phenology, compute the Relative photoperiod effectiveness. The inputs are P=10.46 h, Po=10.5 h, PS=0.095 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF050_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "fraction"} +{"answer": "0.182", "answer_float": 0.182, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.182", "answer_float": 0.182, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Plant methane transport factor (植株传输因子).\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.182", "answer_float": 0.182, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.182", "answer_float": 0.182, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.182", "answer_float": 0.182, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.182", "answer_float": 0.182, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF019", "formula_name_en": "Plant methane transport factor", "id": "CMF019_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Power attenuation", "mode": "direct", "precision": 3, "problem": "In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Plant methane transport factor is computed as part of methane modeling. With CH4RiceEfC=0.217 fraction, W=6048.7 kg/ha, Wmax=11870.1 kg/ha, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF019_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "fraction"} +{"answer": "0.0022", "answer_float": 0.0022, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_chain_2_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "precision": 4, "problem": "Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "d-1"} +{"answer": "0.0022", "answer_float": 0.0022, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_chain_2_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "precision": 4, "problem": "Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "d-1"} +{"answer": "0.0022", "answer_float": 0.0022, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_chain_2_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "precision": 4, "problem": "Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "d-1"} +{"answer": "0.0022", "answer_float": 0.0022, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_chain_2_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "precision": 4, "problem": "Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "d-1"} +{"answer": "0.0022", "answer_float": 0.0022, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_chain_2_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "precision": 4, "problem": "Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "d-1"} +{"answer": "0.0022", "answer_float": 0.0022, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_chain_2_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Minimum stress multiplier", "mode": "chain_2", "precision": 4, "problem": "Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given RP=0.0038 d-1, NNI=0.58 fraction, WDF=0.73 fraction, compute Actual LAI relative growth rate in leaf area growth. This requires sequential calculation through intermediate variables. Report the final numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_chain_2_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "d-1"} +{"answer": "24.14", "answer_float": 24.14, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_chain_2_t1_0001__C", "is_gold": false, "level": 4, "math_type": "Product with Q10 temperature response", "mode": "chain_2", "precision": 2, "problem": "Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "24.14", "answer_float": 24.14, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_chain_2_t1_0001__K_name", "is_gold": false, "level": 4, "math_type": "Product with Q10 temperature response", "mode": "chain_2", "precision": 2, "problem": "Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "24.14", "answer_float": 24.14, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_chain_2_t1_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Product with Q10 temperature response", "mode": "chain_2", "precision": 2, "problem": "Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "24.14", "answer_float": 24.14, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_chain_2_t1_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Product with Q10 temperature response", "mode": "chain_2", "precision": 2, "problem": "Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "24.14", "answer_float": 24.14, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_chain_2_t1_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Product with Q10 temperature response", "mode": "chain_2", "precision": 2, "problem": "Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "24.14", "answer_float": 24.14, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_chain_2_t1_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Product with Q10 temperature response", "mode": "chain_2", "precision": 2, "problem": "Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=55.4 physiological days, ABIOMASS=5647.0 kg/ha, Q10=1.88 dimensionless, T=32.4 C, To=29.5 C, compute Maintenance respiration in respiration. This requires sequential calculation through intermediate variables. Report the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "243.95", "answer_float": 243.95, "change": {"new_value": 187.7, "old_value": 170.6, "parameter": "SLAc"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t0_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "cm2/g"} +{"answer": "243.95", "answer_float": 243.95, "change": {"new_value": 187.7, "old_value": 170.6, "parameter": "SLAc"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t0_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Specific leaf area (比叶面积).\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "cm2/g"} +{"answer": "243.95", "answer_float": 243.95, "change": {"new_value": 187.7, "old_value": 170.6, "parameter": "SLAc"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "cm2/g"} +{"answer": "243.95", "answer_float": 243.95, "change": {"new_value": 187.7, "old_value": 170.6, "parameter": "SLAc"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "cm2/g"} +{"answer": "243.95", "answer_float": 243.95, "change": {"new_value": 187.7, "old_value": 170.6, "parameter": "SLAc"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "cm2/g"} +{"answer": "243.95", "answer_float": 243.95, "change": {"new_value": 187.7, "old_value": 170.6, "parameter": "SLAc"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF053", "formula_name_en": "Specific leaf area", "id": "CMF053_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise polynomial", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are GDD=488.0 C day, SLAc=170.6 cm2/g. A field technician asks what the specific leaf area would become if SLAc changed to 187.7. Report the new final numeric value in cm2/g.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF053_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "cm2/g"} +{"answer": "0.025", "answer_float": 0.025, "change": {"new_value": 0.53, "old_value": 0.48, "parameter": "S_Water"}, "condition": "C", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "response factor"} +{"answer": "0.025", "answer_float": 0.025, "change": {"new_value": 0.53, "old_value": 0.48, "parameter": "S_Water"}, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response water stress response (环境响应水分响应).\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "response factor"} +{"answer": "0.025", "answer_float": 0.025, "change": {"new_value": 0.53, "old_value": 0.48, "parameter": "S_Water"}, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "response factor"} +{"answer": "0.025", "answer_float": 0.025, "change": {"new_value": 0.53, "old_value": 0.48, "parameter": "S_Water"}, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "response factor"} +{"answer": "0.025", "answer_float": 0.025, "change": {"new_value": 0.53, "old_value": 0.48, "parameter": "S_Water"}, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "response factor"} +{"answer": "0.025", "answer_float": 0.025, "change": {"new_value": 0.53, "old_value": 0.48, "parameter": "S_Water"}, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF061", "formula_name_en": "environmental-response water stress response", "id": "CMF061_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 3, "problem": "A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation involves sensitivity analysis. The current measurements are ARID=1.84 dimensionless, S_Water=0.48 dimensionless. A field technician asks what the simple water stress response would become if S_Water changed to 0.53. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF061_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "response factor"} +{"answer": "0.936", "answer_float": 0.936, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.936", "answer_float": 0.936, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.936", "answer_float": 0.936, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.936", "answer_float": 0.936, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.936", "answer_float": 0.936, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.936", "answer_float": 0.936, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Exponential quadratic", "mode": "direct", "precision": 3, "problem": "Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given WaterC=0.455 fraction, calculate the Water-content decomposition factor for a methane. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.094", "answer_float": 0.094, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_branch_t1_0001__C", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.094", "answer_float": 0.094, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_branch_t1_0001__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.094", "answer_float": 0.094, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_branch_t1_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.094", "answer_float": 0.094, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_branch_t1_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.094", "answer_float": 0.094, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_branch_t1_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.094", "answer_float": 0.094, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_branch_t1_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling nitrogen, compute the carbon-nitrogen cycling crop nitrogen demand. The inputs are N_shoot=5.0 g N/m2, Coef_Plant_TN_uptake=0.204 d-1, N_requirement_max_day=2.99 g N/m2/day, NPP=8.58 g C/m2/day, Ratio_N_C_photo_product=0.011 g N/g C, DVI=1.28 dimensionless. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "1.286", "answer_float": 1.286, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "multiplier"} +{"answer": "1.286", "answer_float": 1.286, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "multiplier"} +{"answer": "1.286", "answer_float": 1.286, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "multiplier"} +{"answer": "1.286", "answer_float": 1.286, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "multiplier"} +{"answer": "1.286", "answer_float": 1.286, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "multiplier"} +{"answer": "1.286", "answer_float": 1.286, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response CO2 response. The measured inputs are CO2=557.0 ppm, CO2_RUE=0.138 fraction. What is the simple co2 response? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "multiplier"} +{"answer": "0.0082", "answer_float": 0.0082, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "d-1"} +{"answer": "0.0082", "answer_float": 0.0082, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "d-1"} +{"answer": "0.0082", "answer_float": 0.0082, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "d-1"} +{"answer": "0.0082", "answer_float": 0.0082, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "d-1"} +{"answer": "0.0082", "answer_float": 0.0082, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "d-1"} +{"answer": "0.0082", "answer_float": 0.0082, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=9.3 physiological days, calculate the Maintenance respiration coefficient at optimum temperature for a respiration. The result should be in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "d-1"} +{"answer": "1.02", "answer_float": 1.02, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.02", "answer_float": 1.02, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Leaf area index from SLA (比叶面积法叶面积指数).\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.02", "answer_float": 1.02, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.02", "answer_float": 1.02, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.02", "answer_float": 1.02, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "1.02", "answer_float": 1.02, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF038", "formula_name_en": "Leaf area index from SLA", "id": "CMF038_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Product", "mode": "direct", "precision": 2, "problem": "Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given SLA=0.00037 ha/kg, AWL=2747.0 kg/ha, calculate the Leaf area index from SLA for a leaf area growth. The result should be in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF038_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 1, "unit": "m2/m2"} +{"answer": "0.592", "answer_float": 0.592, "change": {"new_value": -103.8, "old_value": -90.3, "parameter": "Eh"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t1_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.592", "answer_float": 0.592, "change": {"new_value": -103.8, "old_value": -90.3, "parameter": "Eh"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t1_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Redox potential factor (氧化还原电位因子).\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.592", "answer_float": 0.592, "change": {"new_value": -103.8, "old_value": -90.3, "parameter": "Eh"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.592", "answer_float": 0.592, "change": {"new_value": -103.8, "old_value": -90.3, "parameter": "Eh"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.592", "answer_float": 0.592, "change": {"new_value": -103.8, "old_value": -90.3, "parameter": "Eh"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "0.592", "answer_float": 0.592, "change": {"new_value": -103.8, "old_value": -90.3, "parameter": "Eh"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF018", "formula_name_en": "Redox potential factor", "id": "CMF018_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Eh=-90.3 mV in methane, what would the Redox potential factor be if Eh were adjusted to -103.8? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF018_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "fraction"} +{"answer": "2.409", "answer_float": 2.409, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "2.409", "answer_float": 2.409, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "2.409", "answer_float": 2.409, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "2.409", "answer_float": 2.409, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "2.409", "answer_float": 2.409, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "2.409", "answer_float": 2.409, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_chain_2_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Maximum with chained drivers", "mode": "chain_2", "precision": 3, "problem": "In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, the Methane production is obtained by chaining intermediate calculations. With inputs FEh=0.354 fraction, TI=0.289 fraction, Cr=63.36 g C/m2/day, Com=6.896 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 9000.0, "old_value": 9991.0, "parameter": "B_above"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 9000.0, "old_value": 9991.0, "parameter": "B_above"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Harvest index (收获指数).\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 9000.0, "old_value": 9991.0, "parameter": "B_above"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 9000.0, "old_value": 9991.0, "parameter": "B_above"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 9000.0, "old_value": 9991.0, "parameter": "B_above"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 9000.0, "old_value": 9991.0, "parameter": "B_above"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF034", "formula_name_en": "Harvest index", "id": "CMF034_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Harvest index responds to parameter changes. Baseline: Y_grain=5253.0 kg/ha, B_above=9991.0 kg/ha. After changing B_above from 9991.0 to 9000.0, compute the new Harvest index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF034_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.961", "answer_float": 0.961, "change": {"new_value": 1.4, "old_value": 1.27, "parameter": "BD"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_sensitivity_t0_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.961", "answer_float": 0.961, "change": {"new_value": 1.4, "old_value": 1.27, "parameter": "BD"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_sensitivity_t0_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子).\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.961", "answer_float": 0.961, "change": {"new_value": 1.4, "old_value": 1.27, "parameter": "BD"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.961", "answer_float": 0.961, "change": {"new_value": 1.4, "old_value": 1.27, "parameter": "BD"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.961", "answer_float": 0.961, "change": {"new_value": 1.4, "old_value": 1.27, "parameter": "BD"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "0.961", "answer_float": 0.961, "change": {"new_value": 1.4, "old_value": 1.27, "parameter": "BD"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF012", "formula_name_en": "carbon-nitrogen cycling soil carbon water factor", "id": "CMF012_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise exponential", "mode": "sensitivity", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon calculation involves sensitivity analysis. The current measurements are Soil_Water=0.411 cm3/cm3, Saturate_Water=0.48 cm3/cm3, BD=1.27 g/cm3. A field technician asks what the carbon-nitrogen cycling soil carbon water factor would become if BD changed to 1.4. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF012_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "fraction"} +{"answer": "18.30", "answer_float": 18.3, "change": {"new_value": 30.0, "old_value": 33.3, "parameter": "Tmean"}, "condition": "C", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "18.30", "answer_float": 18.3, "change": {"new_value": 30.0, "old_value": 33.3, "parameter": "Tmean"}, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "18.30", "answer_float": 18.3, "change": {"new_value": 30.0, "old_value": 33.3, "parameter": "Tmean"}, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "18.30", "answer_float": 18.3, "change": {"new_value": 30.0, "old_value": 33.3, "parameter": "Tmean"}, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "18.30", "answer_float": 18.3, "change": {"new_value": 30.0, "old_value": 33.3, "parameter": "Tmean"}, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "18.30", "answer_float": 18.3, "change": {"new_value": 30.0, "old_value": 33.3, "parameter": "Tmean"}, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Maximum linear", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=33.3 C, Tbase=11.7 C in environmental-response formula system, what would the environmental-response daily thermal time be if Tmean were adjusted to 30.0? Provide the numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "C day"} +{"answer": "487.96", "answer_float": 487.96, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t0_0000__C", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "487.96", "answer_float": 487.96, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t0_0000__K_name", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "487.96", "answer_float": 487.96, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t0_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "487.96", "answer_float": 487.96, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t0_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "487.96", "answer_float": 487.96, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t0_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "487.96", "answer_float": 487.96, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t0_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling canopy absorbed PAR using intermediate drivers. Use the provided inputs in sequence. The measurements are Radi=485.8 W/m2, LAI=2.51 m2/m2, Coef_rou=0.23 fraction, Coef_K=0.38 dimensionless. Compute the carbon-nitrogen cycling canopy absorbed par and give the final numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol/m2/s"} +{"answer": "0.094", "answer_float": 0.094, "change": {"new_value": 42.0, "old_value": 43.8, "parameter": "ExtremeT"}, "condition": "C", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.094", "answer_float": 0.094, "change": {"new_value": 42.0, "old_value": 43.8, "parameter": "ExtremeT"}, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response heat stress response (环境响应高温响应).\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.094", "answer_float": 0.094, "change": {"new_value": 42.0, "old_value": 43.8, "parameter": "ExtremeT"}, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.094", "answer_float": 0.094, "change": {"new_value": 42.0, "old_value": 43.8, "parameter": "ExtremeT"}, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.094", "answer_float": 0.094, "change": {"new_value": 42.0, "old_value": 43.8, "parameter": "ExtremeT"}, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response heat stress response (环境响应高温响应)\nFormula (LaTeX): F_H=1,\\;0,\\;\\text{or}\\;1-\\frac{T_{max}-MaxT}{ExtremeT-MaxT}\nFormula: F_Heat = if Tmax<=MaxT: 1; if Tmax>ExtremeT: 0; else: max(1-(Tmax-MaxT)/(ExtremeT-MaxT), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmax: daily maximum temperature; unit=C; typical range=[24.0, 46.0]\n- MaxT: heat-stress onset temperature; unit=C; typical range=[32.0, 36.0]\n- ExtremeT: complete heat-stress temperature; unit=C; typical range=[42.0, 48.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "0.094", "answer_float": 0.094, "change": {"new_value": 42.0, "old_value": 43.8, "parameter": "ExtremeT"}, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF058", "formula_name_en": "environmental-response heat stress response", "id": "CMF058_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise linear stress", "mode": "sensitivity", "precision": 3, "problem": "In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In environmental-response formula system, we analyze how environmental-response heat stress response responds to parameter changes. Baseline: Tmax=41.2 C, MaxT=33.5 C, ExtremeT=43.8 C. After changing ExtremeT from 43.8 to 42.0, compute the new environmental-response heat stress response. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF058_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 22.9, "old_value": 19.9, "parameter": "Tmean"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t0_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 22.9, "old_value": 19.9, "parameter": "Tmean"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t0_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 22.9, "old_value": 19.9, "parameter": "Tmean"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 22.9, "old_value": 19.9, "parameter": "Tmean"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 22.9, "old_value": 19.9, "parameter": "Tmean"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 22.9, "old_value": 19.9, "parameter": "Tmean"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation involves sensitivity analysis. The current measurements are PDT=37.4 physiological days, Tmean=19.9 C, Qt=17.4 C day. A field technician asks what the low-temperature fertility factor would become if Tmean changed to 22.9. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 23.0, "old_value": 22.0, "parameter": "Tmean"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 23.0, "old_value": 22.0, "parameter": "Tmean"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 23.0, "old_value": 22.0, "parameter": "Tmean"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 23.0, "old_value": 22.0, "parameter": "Tmean"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 23.0, "old_value": 22.0, "parameter": "Tmean"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 23.0, "old_value": 22.0, "parameter": "Tmean"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise power penalty", "mode": "sensitivity", "precision": 2, "problem": "In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature stress, we analyze how Low-temperature fertility factor responds to parameter changes. Baseline: PDT=33.6 physiological days, Tmean=22.0 C, Qt=3.6 C day. After changing Tmean from 22.0 to 23.0, compute the new Low-temperature fertility factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 2, "unit": "fraction"} +{"answer": "68.00", "answer_float": 68.0, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "68.00", "answer_float": 68.0, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "68.00", "answer_float": 68.0, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "68.00", "answer_float": 68.0, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "68.00", "answer_float": 68.0, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "68.00", "answer_float": 68.0, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration is computed as part of respiration modeling. With PDT=20.6 physiological days, ABIOMASS=8842.0 kg/ha, Q10=1.88 dimensionless, T=30.6 C, To=29.2 C, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "0.576", "answer_float": 0.576, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t1_0001__C", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.576", "answer_float": 0.576, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t1_0001__K_name", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.576", "answer_float": 0.576, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t1_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.576", "answer_float": 0.576, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t1_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.576", "answer_float": 0.576, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t1_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.576", "answer_float": 0.576, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_chain_2_t1_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "chain_2", "precision": 3, "problem": "Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Leaf area index from SLA (比叶面积法叶面积指数)\nFormula (LaTeX): LAI=SLA\\cdot AWL\nFormula: LAI = SLA * AWL\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- SLA: specific leaf area; unit=ha/kg; typical range=[0.00012, 0.00042]\n- AWL: actual green leaf dry weight; unit=kg/ha; typical range=[800.0, 8500.0]\nPhysiological category: Leaf area growth\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=33.7 physiological days, DTE=0.86 fraction, IE=1.09 dimensionless, RPE=0.59 fraction, BFF=0.67 dimensionless, compute Daily physiological effectiveness in phenology. This requires sequential calculation through intermediate variables. Report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_chain_2_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "1.240", "answer_float": 1.24, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0000__C", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.240", "answer_float": 1.24, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0000__K_name", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.240", "answer_float": 1.24, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.240", "answer_float": 1.24, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.240", "answer_float": 1.24, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.240", "answer_float": 1.24, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=550.0 ppm, Coef_B_CO2=0.5 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.339 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "7.802", "answer_float": 7.802, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "7.802", "answer_float": 7.802, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "7.802", "answer_float": 7.802, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "7.802", "answer_float": 7.802, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "7.802", "answer_float": 7.802, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "7.802", "answer_float": 7.802, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis module estimates carbon-nitrogen cycling gross primary production using intermediate drivers. Use the provided inputs in sequence. The measurements are Photo_Rate_opt=46.58 umol CO2/m2/s, F_T=0.84 fraction, F_Water=0.5 fraction, F_CO2=1.02 fraction, Day_Length_hour=9.05 h. Compute the carbon-nitrogen cycling gross primary production and give the final numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "g C/m2/day"} +{"answer": "0.678", "answer_float": 0.678, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.678", "answer_float": 0.678, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.678", "answer_float": 0.678, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.678", "answer_float": 0.678, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.678", "answer_float": 0.678, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.678", "answer_float": 0.678, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Q10 with DVI correction", "mode": "direct", "precision": 3, "problem": "Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Biom_plant_yesterday=691.1 g C/m2, Coef_Rm=0.0013 g C/g C/s, Tmn=23.0 C, DVI=1.63 dimensionless, calculate the carbon-nitrogen cycling plant maintenance respiration for a carbon-nitrogen cycling respiration. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "4.020", "answer_float": 4.02, "change": {"new_value": 39.5, "old_value": 35.91, "parameter": "Photo_Rate_opt"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "4.020", "answer_float": 4.02, "change": {"new_value": 39.5, "old_value": 35.91, "parameter": "Photo_Rate_opt"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "4.020", "answer_float": 4.02, "change": {"new_value": 39.5, "old_value": 35.91, "parameter": "Photo_Rate_opt"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "4.020", "answer_float": 4.02, "change": {"new_value": 39.5, "old_value": 35.91, "parameter": "Photo_Rate_opt"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "4.020", "answer_float": 4.02, "change": {"new_value": 39.5, "old_value": 35.91, "parameter": "Photo_Rate_opt"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "4.020", "answer_float": 4.02, "change": {"new_value": 39.5, "old_value": 35.91, "parameter": "Photo_Rate_opt"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Multiplicative conversion", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, we analyze how carbon-nitrogen cycling gross primary production responds to parameter changes. Baseline: Photo_Rate_opt=35.91 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.39 fraction, F_CO2=0.97 fraction, Day_Length_hour=12.71 h. After changing Photo_Rate_opt from 35.91 to 39.5, compute the new carbon-nitrogen cycling gross primary production. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "6.640", "answer_float": 6.64, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0001__C", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "6.640", "answer_float": 6.64, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0001__K_name", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "6.640", "answer_float": 6.64, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "6.640", "answer_float": 6.64, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "6.640", "answer_float": 6.64, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "6.640", "answer_float": 6.64, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_chain_2_t2_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Multiplicative conversion", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling gross primary production is obtained by chaining intermediate calculations. With inputs Photo_Rate_opt=32.1 umol CO2/m2/s, F_T=0.54 fraction, F_Water=0.54 fraction, F_CO2=1.23 fraction, Day_Length_hour=13.35 h, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.028", "answer_float": 0.028, "change": {"new_value": 1.8, "old_value": 1.65, "parameter": "DVI"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.028", "answer_float": 0.028, "change": {"new_value": 1.8, "old_value": 1.65, "parameter": "DVI"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸).\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.028", "answer_float": 0.028, "change": {"new_value": 1.8, "old_value": 1.65, "parameter": "DVI"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.028", "answer_float": 0.028, "change": {"new_value": 1.8, "old_value": 1.65, "parameter": "DVI"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.028", "answer_float": 0.028, "change": {"new_value": 1.8, "old_value": 1.65, "parameter": "DVI"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.028", "answer_float": 0.028, "change": {"new_value": 1.8, "old_value": 1.65, "parameter": "DVI"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF008", "formula_name_en": "carbon-nitrogen cycling plant maintenance respiration", "id": "CMF008_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Q10 with DVI correction", "mode": "sensitivity", "precision": 3, "problem": "In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling respiration, we analyze how carbon-nitrogen cycling plant maintenance respiration responds to parameter changes. Baseline: Biom_plant_yesterday=131.0 g C/m2, Coef_Rm=0.0005 g C/g C/s, Tmn=17.4 C, DVI=1.65 dimensionless. After changing DVI from 1.65 to 1.8, compute the new carbon-nitrogen cycling plant maintenance respiration. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF008_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling respiration", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "10.44", "answer_float": 10.44, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "10.44", "answer_float": 10.44, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual maximum assimilation rate (实际最大同化速率).\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "10.44", "answer_float": 10.44, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "10.44", "answer_float": 10.44, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "10.44", "answer_float": 10.44, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: High-temperature fertility factor (高温结实因子)\nFormula (LaTeX): HTF=\\frac{1}{1+e^{0.853(T_m-36.6)}}\nFormula: HTF = if 32<=PDT<=39 and Tmax_day>36.6: 1/(1+exp(0.853*(Tmax_day-36.6))); else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[32.0, 39.0]\n- Tmax_day: daily maximum temperature; unit=C; typical range=[34.0, 42.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "10.44", "answer_float": 10.44, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF026", "formula_name_en": "Actual maximum assimilation rate", "id": "CMF026_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative with minimum stress", "mode": "direct", "precision": 2, "problem": "Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Am=53.7 kg CO2/ha/h, FCO2=1.27 fraction, FPA=0.52 fraction, FT=0.64 fraction, NNI=0.46 fraction, WDF=0.89 fraction, calculate the Actual maximum assimilation rate for a photosynthesis. The result should be in kg CO2/ha/h.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF026_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "kg CO2/ha/h"} +{"answer": "18.10", "answer_float": 18.1, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0000__C", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "18.10", "answer_float": 18.1, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0000__K_name", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "18.10", "answer_float": 18.1, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "18.10", "answer_float": 18.1, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "18.10", "answer_float": 18.1, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "18.10", "answer_float": 18.1, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=29.6 C, Tbase=11.5 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "8.46", "answer_float": 8.46, "change": {"new_value": 0.57, "old_value": 0.52, "parameter": "k"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Product with exponential saturation", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "8.46", "answer_float": 8.46, "change": {"new_value": 0.57, "old_value": 0.52, "parameter": "k"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Product with exponential saturation", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Intercepted photosynthetically active radiation (截获光合有效辐射).\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "8.46", "answer_float": 8.46, "change": {"new_value": 0.57, "old_value": 0.52, "parameter": "k"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Product with exponential saturation", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "8.46", "answer_float": 8.46, "change": {"new_value": 0.57, "old_value": 0.52, "parameter": "k"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Product with exponential saturation", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "8.46", "answer_float": 8.46, "change": {"new_value": 0.57, "old_value": 0.52, "parameter": "k"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Product with exponential saturation", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "8.46", "answer_float": 8.46, "change": {"new_value": 0.57, "old_value": 0.52, "parameter": "k"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF047", "formula_name_en": "Intercepted photosynthetically active radiation", "id": "CMF047_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Product with exponential saturation", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Harvest index (收获指数)\nFormula (LaTeX): HI=\\frac{Y_{grain}}{B_{above}}\nFormula: HI = Y_grain / B_above\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PAR=12.51 MJ/m2/day, k=0.52 dimensionless, LAI=1.98 m2/m2. A field technician asks what the intercepted photosynthetically active radiation would become if k changed to 0.57. Report the new final numeric value in MJ/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF047_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "MJ/m2/day"} +{"answer": "20.32", "answer_float": 20.32, "change": {"new_value": 120.0, "old_value": 113.3, "parameter": "RM"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Subtraction and conversion", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "20.32", "answer_float": 20.32, "change": {"new_value": 120.0, "old_value": 113.3, "parameter": "RM"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Subtraction and conversion", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "20.32", "answer_float": 20.32, "change": {"new_value": 120.0, "old_value": 113.3, "parameter": "RM"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Subtraction and conversion", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "20.32", "answer_float": 20.32, "change": {"new_value": 120.0, "old_value": 113.3, "parameter": "RM"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Subtraction and conversion", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "20.32", "answer_float": 20.32, "change": {"new_value": 120.0, "old_value": 113.3, "parameter": "RM"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Subtraction and conversion", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative photoperiod effectiveness (相对光周期效应)\nFormula (LaTeX): RPE=1,\\;1-PS(P-P_o)^2,\\;0\nFormula: RPE = if P<=Po: 1; if P<=Po+sqrt(1/PS): 1-PS*(P-Po)^2; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- P: observed daylength; unit=h; typical range=[9.5, 15.5]\n- Po: optimum daylength; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Piecewise quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "20.32", "answer_float": 20.32, "change": {"new_value": 120.0, "old_value": 113.3, "parameter": "RM"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Subtraction and conversion", "mode": "sensitivity", "precision": 2, "problem": "In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, we analyze how Daily dry matter increment responds to parameter changes. Baseline: DTGA=194.1 kg/ha/day, RM=113.3 kg/ha/day, RG=54.8 kg/ha/day, b=0.05 fraction. After changing RM from 113.3 to 120.0, compute the new Daily dry matter increment. Give the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "0.040", "answer_float": 0.04, "change": {"new_value": 0.257, "old_value": 0.285, "parameter": "WI"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_sensitivity_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.040", "answer_float": 0.04, "change": {"new_value": 0.257, "old_value": 0.285, "parameter": "WI"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_sensitivity_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.040", "answer_float": 0.04, "change": {"new_value": 0.257, "old_value": 0.285, "parameter": "WI"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.040", "answer_float": 0.04, "change": {"new_value": 0.257, "old_value": 0.285, "parameter": "WI"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.040", "answer_float": 0.04, "change": {"new_value": 0.257, "old_value": 0.285, "parameter": "WI"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Redox potential factor (氧化还原电位因子)\nFormula (LaTeX): F_{Eh}=1\\;(Eh<-150),\\quad e^{-1.7(1+Eh/150)}\nFormula: FEh = if Eh<-150: 1; else: exp(-1.7*(1+Eh/150))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Eh: soil redox potential; unit=mV; typical range=[-230.0, 160.0]\nPhysiological category: Methane\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.040", "answer_float": 0.04, "change": {"new_value": 0.257, "old_value": 0.285, "parameter": "WI"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Soil organic matter carbon decomposition responds to parameter changes. Baseline: WI=0.285 fraction, SI=0.69 fraction, TI=0.25 fraction, OMS=298.4 g C/m2. After changing WI from 0.285 to 0.257, compute the new Soil organic matter carbon decomposition. Give the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 32.6, "old_value": 29.6, "parameter": "T"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 32.6, "old_value": 29.6, "parameter": "T"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 32.6, "old_value": 29.6, "parameter": "T"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 32.6, "old_value": 29.6, "parameter": "T"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 32.6, "old_value": 29.6, "parameter": "T"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 32.6, "old_value": 29.6, "parameter": "T"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Emergence thermal requirement (出苗热量需求)\nFormula (LaTeX): EMGDD=45+7SDEPTH\nFormula: EMGDD = 45 + 7 * SDEPTH\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- SDEPTH: sowing depth; unit=cm; typical range=[1.0, 5.5]\nPhysiological category: Phenology\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements T=29.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 32.6? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "1.15", "answer_float": 1.15, "change": {"new_value": 0.0047, "old_value": 0.0041, "parameter": "RP"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Exponential growth", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "1.15", "answer_float": 1.15, "change": {"new_value": 0.0047, "old_value": 0.0041, "parameter": "RP"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Exponential growth", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Potential leaf area index (潜在叶面积指数).\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "1.15", "answer_float": 1.15, "change": {"new_value": 0.0047, "old_value": 0.0041, "parameter": "RP"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Exponential growth", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "1.15", "answer_float": 1.15, "change": {"new_value": 0.0047, "old_value": 0.0041, "parameter": "RP"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Exponential growth", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "1.15", "answer_float": 1.15, "change": {"new_value": 0.0047, "old_value": 0.0041, "parameter": "RP"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Exponential growth", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "1.15", "answer_float": 1.15, "change": {"new_value": 0.0047, "old_value": 0.0041, "parameter": "RP"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF037", "formula_name_en": "Potential leaf area index", "id": "CMF037_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Exponential growth", "mode": "sensitivity", "precision": 2, "problem": "A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A leaf area growth calculation involves sensitivity analysis. The current measurements are LAI0=0.315 m2/m2, RP=0.0041 d-1, GDD=275.6 C day. A field technician asks what the potential leaf area index would become if RP changed to 0.0047. Report the new final numeric value in m2/m2.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF037_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 0, "unit": "m2/m2"} +{"answer": "10.60", "answer_float": 10.6, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "10.60", "answer_float": 10.6, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily thermal time (每日有效积温).\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "10.60", "answer_float": 10.6, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "10.60", "answer_float": 10.6, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "10.60", "answer_float": 10.6, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "10.60", "answer_float": 10.6, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF056", "formula_name_en": "environmental-response daily thermal time", "id": "CMF056_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Maximum linear", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily thermal time. The measured inputs are Tmean=15.7 C, Tbase=5.1 C. What is the simple daily thermal time? Give the final numeric value in C day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF056_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "C day"} +{"answer": "0.44", "answer_float": 0.44, "change": {"new_value": 22.4, "old_value": 20.4, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF041", "formula_name_en": "Green leaf partitioning index", "id": "CMF041_sensitivity_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF041_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.44", "answer_float": 0.44, "change": {"new_value": 22.4, "old_value": 20.4, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF041", "formula_name_en": "Green leaf partitioning index", "id": "CMF041_sensitivity_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Green leaf partitioning index (绿叶分配指数).\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF041_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.44", "answer_float": 0.44, "change": {"new_value": 22.4, "old_value": 20.4, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF041", "formula_name_en": "Green leaf partitioning index", "id": "CMF041_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF041_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.44", "answer_float": 0.44, "change": {"new_value": 22.4, "old_value": 20.4, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF041", "formula_name_en": "Green leaf partitioning index", "id": "CMF041_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF041_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.44", "answer_float": 0.44, "change": {"new_value": 22.4, "old_value": 20.4, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF041", "formula_name_en": "Green leaf partitioning index", "id": "CMF041_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Light interception fraction (光能截获率)\nFormula (LaTeX): fPAR = 1 - e^{-k \\cdot LAI}\nFormula: fPAR = 1 - exp(-k * LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF041_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.44", "answer_float": 0.44, "change": {"new_value": 22.4, "old_value": 20.4, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF041", "formula_name_en": "Green leaf partitioning index", "id": "CMF041_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise linear/exponential", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Green leaf partitioning index responds to parameter changes. Baseline: PDT=20.4 physiological days. After changing PDT from 20.4 to 22.4, compute the new Green leaf partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF041_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.563", "answer_float": 0.563, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t2_0000__C", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.563", "answer_float": 0.563, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t2_0000__K_name", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失).\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.563", "answer_float": 0.563, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t2_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.563", "answer_float": 0.563, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t2_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.563", "answer_float": 0.563, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t2_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.563", "answer_float": 0.563, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF009", "formula_name_en": "carbon-nitrogen cycling carbon pool respiration loss", "id": "CMF009_chain_2_t2_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Chained multiplicative decomposition", "mode": "chain_2", "precision": 3, "problem": "In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling soil carbon, the carbon-nitrogen cycling carbon pool respiration loss is obtained by chaining intermediate calculations. With inputs Pool=514.7 g C/m2, Coef_K_Decom=0.0102 d-1, F_temp=1.209 fraction, F_water=0.532 fraction, F_texture=0.959 fraction, F_pH=0.18 fraction, F_notill=1.14 fraction, Coef_K_Trans=0.152 fraction, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF009_chain_2_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 24.0, "old_value": 21.8, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t1_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 24.0, "old_value": 21.8, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t1_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 24.0, "old_value": 21.8, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 24.0, "old_value": 21.8, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 24.0, "old_value": 21.8, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 24.0, "old_value": 21.8, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements PDT=21.8 physiological days, a=0.039 dimensionless in photosynthesis, what would the Physiological age factor be if PDT were adjusted to 24.0? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 1, "unit": "fraction"} +{"answer": "0.16", "answer_float": 0.16, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.16", "answer_float": 0.16, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.16", "answer_float": 0.16, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.16", "answer_float": 0.16, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.16", "answer_float": 0.16, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.16", "answer_float": 0.16, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Residual from shoot partitioning", "mode": "direct", "precision": 2, "problem": "Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Intercepted photosynthetically active radiation (截获光合有效辐射)\nFormula (LaTeX): IPAR = PAR \\cdot (1 - e^{-k \\cdot LAI})\nFormula: IPAR = PAR * (1 - exp(-k * LAI))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: MJ/m2/day\nParameters:\n- PAR: incoming photosynthetically active radiation; unit=MJ/m2/day; typical range=[6.0, 18.0]\n- k: canopy light extinction coefficient; unit=dimensionless; typical range=[0.35, 0.75]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.8, 6.5]\nPhysiological category: Photosynthesis\nMathematical form: Product with exponential saturation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=27.6 physiological days, calculate the Root partitioning index for a dry matter partitioning. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 1, "unit": "fraction"} +{"answer": "0.343", "answer_float": 0.343, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.343", "answer_float": 0.343, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.343", "answer_float": 0.343, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.343", "answer_float": 0.343, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.343", "answer_float": 0.343, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.343", "answer_float": 0.343, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_direct_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Beta temperature response", "mode": "direct", "precision": 3, "problem": "A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation estimates Relative thermal effectiveness. The measured inputs are T=15.0 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.19 dimensionless. What is the relative thermal effectiveness? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "204.19", "answer_float": 204.19, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol/m2/s"} +{"answer": "204.19", "answer_float": 204.19, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射).\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol/m2/s"} +{"answer": "204.19", "answer_float": 204.19, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol/m2/s"} +{"answer": "204.19", "answer_float": 204.19, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol/m2/s"} +{"answer": "204.19", "answer_float": 204.19, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol/m2/s"} +{"answer": "204.19", "answer_float": 204.19, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF006", "formula_name_en": "carbon-nitrogen cycling canopy absorbed PAR", "id": "CMF006_chain_2_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Exponential ratio", "mode": "chain_2", "precision": 2, "problem": "In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In carbon-nitrogen cycling photosynthesis, the carbon-nitrogen cycling canopy absorbed PAR is obtained by chaining intermediate calculations. With inputs Radi=411.2 W/m2, LAI=5.57 m2/m2, Coef_rou=0.15 fraction, Coef_K=0.6 dimensionless, determine the numeric value in umol/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF006_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 2, "unit": "umol/m2/s"} +{"answer": "0.901", "answer_float": 0.901, "change": {"new_value": 0.566, "old_value": 0.629, "parameter": "TI"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.901", "answer_float": 0.901, "change": {"new_value": 0.566, "old_value": 0.629, "parameter": "TI"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量).\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.901", "answer_float": 0.901, "change": {"new_value": 0.566, "old_value": 0.629, "parameter": "TI"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.901", "answer_float": 0.901, "change": {"new_value": 0.566, "old_value": 0.629, "parameter": "TI"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.901", "answer_float": 0.901, "change": {"new_value": 0.566, "old_value": 0.629, "parameter": "TI"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Soil temperature from air temperature (土壤温度估算)\nFormula (LaTeX): T_{soil}=4.4+0.76T_{air}\nFormula: Tsoil = 4.4 + 0.76 * Tair\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C\nParameters:\n- Tair: air temperature; unit=C; typical range=[12.0, 36.0]\nPhysiological category: Methane\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "0.901", "answer_float": 0.901, "change": {"new_value": 0.566, "old_value": 0.629, "parameter": "TI"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF020", "formula_name_en": "Fresh organic matter carbon decomposition", "id": "CMF020_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements WI=0.26 fraction, SI=0.91 fraction, TI=0.629 fraction, OMN=249.2 g C/m2 in methane, what would the Fresh organic matter carbon decomposition be if TI were adjusted to 0.566? Provide the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF020_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "1.605", "answer_float": 1.605, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "multiplier"} +{"answer": "1.605", "answer_float": 1.605, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "multiplier"} +{"answer": "1.605", "answer_float": 1.605, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "multiplier"} +{"answer": "1.605", "answer_float": 1.605, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "multiplier"} +{"answer": "1.605", "answer_float": 1.605, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "multiplier"} +{"answer": "1.605", "answer_float": 1.605, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In an environmental-response formula system, the environmental-response CO2 response is computed as part of environmental-response formula calculation. With CO2=739.0 ppm, CO2_RUE=0.173 fraction, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 2, "unit": "multiplier"} +{"answer": "6.456", "answer_float": 6.456, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "6.456", "answer_float": 6.456, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling gross primary production (总初级生产力).\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "6.456", "answer_float": 6.456, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "6.456", "answer_float": 6.456, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "6.456", "answer_float": 6.456, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling gross primary production (总初级生产力)\nFormula (LaTeX): GPP=P_{opt}F_TF_WF_{CO2}DL\\cdot0.0432\nFormula: GPP = Photo_Rate_opt * F_T * F_Water * F_CO2 * Day_Length_hour * 0.0432\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Photo_Rate_opt: optimal photosynthetic rate; unit=umol CO2/m2/s; typical range=[5.0, 60.0]\n- F_T: temperature factor; unit=fraction; typical range=[0.3, 1.0]\n- F_Water: water factor; unit=fraction; typical range=[0.3, 1.0]\n- F_CO2: CO2 factor; unit=fraction; typical range=[0.9, 1.35]\n- Day_Length_hour: day length; unit=h; typical range=[9.0, 15.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Multiplicative conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "6.456", "answer_float": 6.456, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF002", "formula_name_en": "carbon-nitrogen cycling gross primary production", "id": "CMF002_direct_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative conversion", "mode": "direct", "precision": 3, "problem": "Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling carbon pool respiration loss (碳库呼吸损失)\nFormula (LaTeX): R=PoolK F_TF_WF_{texture}F_{pH}F_{notill}(1-K_{trans})\nFormula: Respiration = Pool*Coef_K_Decom*F_temp*F_water*F_texture*F_pH*F_notill*(1-Coef_K_Trans)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Pool: carbon pool size; unit=g C/m2; typical range=[120.0, 2200.0]\n- Coef_K_Decom: decomposition rate coefficient; unit=d-1; typical range=[0.0002, 0.012]\n- F_temp: temperature factor; unit=fraction; typical range=[0.4, 2.5]\n- F_water: water factor; unit=fraction; typical range=[0.25, 0.9]\n- F_texture: texture factor; unit=fraction; typical range=[0.75, 1.0]\n- F_pH: pH factor; unit=fraction; typical range=[0.1, 1.0]\n- F_notill: tillage factor; unit=fraction; typical range=[0.75, 1.15]\n- Coef_K_Trans: transport coefficient; unit=fraction; typical range=[0.05, 0.45]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Chained multiplicative decomposition\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given Photo_Rate_opt=51.89 umol CO2/m2/s, F_T=0.49 fraction, F_Water=0.43 fraction, F_CO2=1.14 fraction, Day_Length_hour=11.99 h, calculate the carbon-nitrogen cycling gross primary production for a carbon-nitrogen cycling photosynthesis. The result should be in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF002_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "g C/m2/day"} +{"answer": "113.00", "answer_float": 113.0, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "113.00", "answer_float": 113.0, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response daily biomass increment (环境响应日生物量增量).\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "113.00", "answer_float": 113.0, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "113.00", "answer_float": 113.0, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "113.00", "answer_float": 113.0, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "113.00", "answer_float": 113.0, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF055", "formula_name_en": "environmental-response daily biomass increment", "id": "CMF055_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Multiplicative minimum stress", "mode": "direct", "precision": 2, "problem": "A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response water stress response (环境响应水分响应)\nFormula (LaTeX): F_W=\\max(0,1-S_WARID)\nFormula: F_Water = max(0, 1 - S_Water * ARID)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- ARID: aridity index; unit=dimensionless; typical range=[0.0, 2.5]\n- S_Water: water-stress sensitivity; unit=dimensionless; typical range=[0.15, 0.75]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A environmental-response formula system calculation estimates environmental-response daily biomass increment. The measured inputs are RUE=2.5 g/MJ, fSolar=0.65 fraction, SRAD=12.13 MJ/m2/day, F_CO2=1.26 fraction, F_Temp=0.7 fraction, F_Water=0.65 fraction, F_Heat=1.0 fraction. What is the simple daily biomass increment? Give the final numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF055_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 0, "unit": "kg/ha/day"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 17.8, "old_value": 16.2, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 17.8, "old_value": 16.2, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Physiological age factor (生理年龄因子).\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 17.8, "old_value": 16.2, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 17.8, "old_value": 16.2, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 17.8, "old_value": 16.2, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 17.8, "old_value": 16.2, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF030", "formula_name_en": "Physiological age factor", "id": "CMF030_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise exponential decay", "mode": "sensitivity", "precision": 2, "problem": "In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In photosynthesis, we analyze how Physiological age factor responds to parameter changes. Baseline: PDT=16.2 physiological days, a=0.035 dimensionless. After changing PDT from 16.2 to 17.8, compute the new Physiological age factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF030_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 2, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_direct_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Piecewise power penalty", "mode": "direct", "precision": 2, "problem": "A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_direct_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise power penalty", "mode": "direct", "precision": 2, "problem": "A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Low-temperature fertility factor (低温结实因子).\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_direct_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise power penalty", "mode": "direct", "precision": 2, "problem": "A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_direct_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise power penalty", "mode": "direct", "precision": 2, "problem": "A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_direct_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise power penalty", "mode": "direct", "precision": 2, "problem": "A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF039_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.94", "answer_float": 0.94, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF039", "formula_name_en": "Low-temperature fertility factor", "id": "CMF039_direct_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise power penalty", "mode": "direct", "precision": 2, "problem": "A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A temperature stress calculation estimates Low-temperature fertility factor. The measured inputs are PDT=29.3 physiological days, Tmean=18.4 C, Qt=7.8 C day. What is the low-temperature fertility factor? Give the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF039_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature stress", "template_id": 0, "unit": "fraction"} +{"answer": "0.922", "answer_float": 0.922, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0001__C", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.922", "answer_float": 0.922, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0001__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.922", "answer_float": 0.922, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0001__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.922", "answer_float": 0.922, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0001__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.922", "answer_float": 0.922, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0001__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "0.922", "answer_float": 0.922, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_branch_t1_0001__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise water response", "mode": "branch", "precision": 3, "problem": "For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For carbon-nitrogen cycling water, compute the carbon-nitrogen cycling crop water stress factor. The inputs are Soil_Water=0.511 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_branch_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 1, "unit": "fraction"} +{"answer": "52.19", "answer_float": 52.19, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "52.19", "answer_float": 52.19, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration (维持呼吸).\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "52.19", "answer_float": 52.19, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "52.19", "answer_float": 52.19, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "52.19", "answer_float": 52.19, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "52.19", "answer_float": 52.19, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF048", "formula_name_en": "Maintenance respiration", "id": "CMF048_direct_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Product with Q10 temperature response", "mode": "direct", "precision": 2, "problem": "Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, ABIOMASS=7608.0 kg/ha, Q10=2.17 dimensionless, T=31.8 C, To=28.1 C, calculate the Maintenance respiration for a respiration. The result should be in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF048_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 1, "unit": "kg/ha/day"} +{"answer": "0.32", "answer_float": 0.32, "change": {"new_value": 5.6, "old_value": 4.9, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.32", "answer_float": 0.32, "change": {"new_value": 5.6, "old_value": 4.9, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Root partitioning index (地下部分配指数).\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.32", "answer_float": 0.32, "change": {"new_value": 5.6, "old_value": 4.9, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.32", "answer_float": 0.32, "change": {"new_value": 5.6, "old_value": 4.9, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.32", "answer_float": 0.32, "change": {"new_value": 5.6, "old_value": 4.9, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Critical daylength (临界日长)\nFormula (LaTeX): P_c=P_o+\\sqrt{1/PS}\nFormula: Pc = Po + sqrt(1 / PS)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: h\nParameters:\n- Po: optimum daylength for rice photoperiod response; unit=h; typical range=[10.5, 10.5]\n- PS: photoperiod sensitivity; unit=dimensionless; typical range=[0.08, 0.35]\nPhysiological category: Phenology\nMathematical form: Power response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.32", "answer_float": 0.32, "change": {"new_value": 5.6, "old_value": 4.9, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF043", "formula_name_en": "Root partitioning index", "id": "CMF043_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Residual from shoot partitioning", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Root partitioning index responds to parameter changes. Baseline: PDT=4.9 physiological days. After changing PDT from 4.9 to 5.6, compute the new Root partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF043_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": {"new_value": 25.4, "old_value": 23.1, "parameter": "Topt"}, "condition": "C", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": {"new_value": 25.4, "old_value": 23.1, "parameter": "Topt"}, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response temperature response (环境响应温度响应).\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": {"new_value": 25.4, "old_value": 23.1, "parameter": "Topt"}, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": {"new_value": 25.4, "old_value": 23.1, "parameter": "Topt"}, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": {"new_value": 25.4, "old_value": 23.1, "parameter": "Topt"}, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response temperature response (环境响应温度响应)\nFormula (LaTeX): F_T=1\\;\\text{or}\\;\\max((T-T_b)/(T_o-T_b),0)\nFormula: F_Temp = if Tmean>=Topt: 1; else: max((Tmean-Tbase)/(Topt-Tbase), 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 10.0]\n- Topt: optimum temperature; unit=C; typical range=[22.0, 30.0]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response solar interception fraction (环境响应光截获比例)\nFormula (LaTeX): fSolar=\\min(f_1,f_2)F_{solar,water}\nFormula: fSolar = min(fSolarMax/(1+exp(-0.01*(TT-I50A))), fSolarMax/(1+exp(0.01*(TT-(Tsum-I50B))))) * min(fSolar_water, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- fSolarMax: maximum solar interception; unit=fraction; typical range=[0.75, 0.98]\n- TT: thermal time; unit=C day; typical range=[120.0, 1700.0]\n- I50A: thermal time at half canopy expansion; unit=C day; typical range=[180.0, 520.0]\n- I50B: thermal time at half senescence; unit=C day; typical range=[180.0, 560.0]\n- Tsum: thermal time to maturity; unit=C day; typical range=[1200.0, 2300.0]\n- fSolar_water: water modifier of solar interception; unit=fraction; typical range=[0.6, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Double logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "0.000", "answer_float": 0.0, "change": {"new_value": 25.4, "old_value": 23.1, "parameter": "Topt"}, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF060", "formula_name_en": "environmental-response temperature response", "id": "CMF060_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise linear", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements Tmean=0.2 C, Tbase=4.3 C, Topt=23.1 C in environmental-response formula system, what would the environmental-response temperature response be if Topt were adjusted to 25.4? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF060_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "fraction"} +{"answer": "1.100", "answer_float": 1.1, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "1.100", "answer_float": 1.1, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子).\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "1.100", "answer_float": 1.1, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "1.100", "answer_float": 1.1, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "1.100", "answer_float": 1.1, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "1.100", "answer_float": 1.1, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF011", "formula_name_en": "carbon-nitrogen cycling soil carbon temperature factor", "id": "CMF011_branch_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Q10 temperature response", "mode": "branch", "precision": 3, "problem": "A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling plant maintenance respiration (植株维持呼吸)\nFormula (LaTeX): R_m=0.0432\\cdot24(Biom/0.45)C_m2^{(T-25)/10}\nFormula: Rm = 0.0432*24*(Biom_plant_yesterday/0.45)*Coef_Rm*2^((Tmn-25)/10) / (if DVI>1: DVI^2; else: 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- Biom_plant_yesterday: previous plant biomass; unit=g C/m2; typical range=[80.0, 950.0]\n- Coef_Rm: maintenance respiration coefficient; unit=g C/g C/s; typical range=[0.0004, 0.004]\n- Tmn: daily mean temperature; unit=C; typical range=[8.0, 34.0]\n- DVI: development index; unit=dimensionless; typical range=[0.2, 1.8]\nPhysiological category: carbon-nitrogen cycling respiration\nMathematical form: Q10 with DVI correction\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling soil carbon involves a stage that may fall on either side of a piecewise threshold. The measured inputs are Temp=8.8 C, is_soil_temp=0.0 0/1. Determine the carbon-nitrogen cycling soil carbon temperature factor and report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF011_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling soil carbon", "template_id": 0, "unit": "multiplier"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子).\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling crop water stress factor (作物水分胁迫因子)\nFormula (LaTeX): F_W=f(\\theta,\\theta_{wp},\\theta_{min},\\theta_{max})\nFormula: F_Water = if Soil_Water<=Wilt_point: 0; if Soil_Water\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "1.000", "answer_float": 1.0, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF014", "formula_name_en": "carbon-nitrogen cycling crop water stress factor", "id": "CMF014_direct_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise water response", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling water, the carbon-nitrogen cycling crop water stress factor is computed as part of carbon-nitrogen cycling water modeling. With Soil_Water=0.288 cm3/cm3, Crop_Water_max=0.42 cm3/cm3, Crop_Water_min=0.22 cm3/cm3, Wilt_point=0.08 cm3/cm3, determine the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF014_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling water", "template_id": 2, "unit": "fraction"} +{"answer": "1.266", "answer_float": 1.266, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t1_0000__C", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.266", "answer_float": 1.266, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t1_0000__K_name", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.266", "answer_float": 1.266, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t1_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.266", "answer_float": 1.266, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t1_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.266", "answer_float": 1.266, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t1_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response yield (环境响应产量)\nFormula (LaTeX): Yield=Biomass\\cdot HI\nFormula: Yield = Biomass * HI\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- Biomass: above-ground biomass; unit=kg/ha; typical range=[3500.0, 22000.0]\n- HI: harvest index; unit=fraction; typical range=[0.25, 0.62]\nPhysiological category: environmental-response formula system\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.266", "answer_float": 1.266, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_direct_t1_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Piecewise linear cap", "mode": "direct", "precision": 3, "problem": "Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=630.0 ppm, CO2_RUE=0.095 fraction, calculate the environmental-response CO2 response for an environmental-response formula system. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_direct_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "0.308", "answer_float": 0.308, "change": {"new_value": 40.0, "old_value": 39.5, "parameter": "T"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_sensitivity_t0_0001__C", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.308", "answer_float": 0.308, "change": {"new_value": 40.0, "old_value": 39.5, "parameter": "T"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_sensitivity_t0_0001__K_name", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Relative thermal effectiveness (相对热效应).\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.308", "answer_float": 0.308, "change": {"new_value": 40.0, "old_value": 39.5, "parameter": "T"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.308", "answer_float": 0.308, "change": {"new_value": 40.0, "old_value": 39.5, "parameter": "T"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.308", "answer_float": 0.308, "change": {"new_value": 40.0, "old_value": 39.5, "parameter": "T"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.308", "answer_float": 0.308, "change": {"new_value": 40.0, "old_value": 39.5, "parameter": "T"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF051", "formula_name_en": "Relative thermal effectiveness", "id": "CMF051_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Beta temperature response", "mode": "sensitivity", "precision": 3, "problem": "A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology calculation involves sensitivity analysis. The current measurements are T=39.5 C, Tb=10.0 C, To=30.0 C, Tm=42.0 C, TS=1.76 dimensionless. A field technician asks what the relative thermal effectiveness would become if T changed to 40.0. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF051_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "fraction"} +{"answer": "0.294", "answer_float": 0.294, "change": {"new_value": 8.2, "old_value": 7.1, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "dimensionless"} +{"answer": "0.294", "answer_float": 0.294, "change": {"new_value": 8.2, "old_value": 7.1, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Canopy extinction coefficient (冠层消光系数).\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "dimensionless"} +{"answer": "0.294", "answer_float": 0.294, "change": {"new_value": 8.2, "old_value": 7.1, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "dimensionless"} +{"answer": "0.294", "answer_float": 0.294, "change": {"new_value": 8.2, "old_value": 7.1, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "dimensionless"} +{"answer": "0.294", "answer_float": 0.294, "change": {"new_value": 8.2, "old_value": 7.1, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Canopy extinction coefficient (冠层消光系数)\nFormula (LaTeX): K=0.0087PDT+0.2222\nFormula: K = 0.0087 * PDT + 0.2222\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: dimensionless\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Photosynthesis\nMathematical form: Linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Specific leaf area (比叶面积)\nFormula (LaTeX): SLA=(-0.0002GDD^2-0.5604GDD+581.04)(SLA_c/200)\nFormula: SLA = if GDD<=1200: (-0.0002*GDD^2-0.5604*GDD+581.04)*(SLAc/200); else: 200*(SLAc/200)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: cm2/g\nParameters:\n- GDD: growing degree days; unit=C day; typical range=[50.0, 800.0]\n- SLAc: cultivar specific leaf area parameter; unit=cm2/g; typical range=[140.0, 260.0]\nPhysiological category: Leaf area growth\nMathematical form: Piecewise polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "dimensionless"} +{"answer": "0.294", "answer_float": 0.294, "change": {"new_value": 8.2, "old_value": 7.1, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF036", "formula_name_en": "Canopy extinction coefficient", "id": "CMF036_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Linear", "mode": "sensitivity", "precision": 3, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Growth respiration (生长呼吸)\nFormula (LaTeX): RG=R_g\\cdot DTGA\nFormula: RG = Rg * DTGA\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- Rg: growth respiration coefficient; unit=fraction; typical range=[0.18, 0.42]\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[120.0, 520.0]\nPhysiological category: Respiration\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are PDT=7.1 physiological days. A field technician asks what the canopy extinction coefficient would become if PDT changed to 8.2. Report the new final numeric value in dimensionless.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF036_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "dimensionless"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 33.8, "old_value": 37.6, "parameter": "T"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 33.8, "old_value": 37.6, "parameter": "T"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 33.8, "old_value": 37.6, "parameter": "T"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 33.8, "old_value": 37.6, "parameter": "T"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 33.8, "old_value": 37.6, "parameter": "T"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "1.00", "answer_float": 1.0, "change": {"new_value": 33.8, "old_value": 37.6, "parameter": "T"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In temperature response, we analyze how Temperature response factor responds to parameter changes. Baseline: T=37.6 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C. After changing T from 37.6 to 33.8, compute the new Temperature response factor. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 2, "unit": "fraction"} +{"answer": "0.0046", "answer_float": 0.0046, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Minimum stress multiplier", "mode": "direct", "precision": 4, "problem": "In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0046", "answer_float": 0.0046, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Minimum stress multiplier", "mode": "direct", "precision": 4, "problem": "In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Actual LAI relative growth rate (实际叶面积相对增长率).\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0046", "answer_float": 0.0046, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Minimum stress multiplier", "mode": "direct", "precision": 4, "problem": "In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0046", "answer_float": 0.0046, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Minimum stress multiplier", "mode": "direct", "precision": 4, "problem": "In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0046", "answer_float": 0.0046, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Minimum stress multiplier", "mode": "direct", "precision": 4, "problem": "In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual LAI relative growth rate (实际叶面积相对增长率)\nFormula (LaTeX): R=RP\\cdot\\min(NNI,WDF)\nFormula: R = RP * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- RP: potential relative growth rate; unit=d-1; typical range=[0.003, 0.009]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Leaf area growth\nMathematical form: Minimum stress multiplier\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "0.0046", "answer_float": 0.0046, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF052", "formula_name_en": "Actual LAI relative growth rate", "id": "CMF052_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Minimum stress multiplier", "mode": "direct", "precision": 4, "problem": "In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a leaf area growth, the Actual LAI relative growth rate is computed as part of leaf area growth modeling. With RP=0.0058 d-1, NNI=0.79 fraction, WDF=0.96 fraction, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF052_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Leaf area growth", "template_id": 2, "unit": "d-1"} +{"answer": "398.53", "answer_float": 398.53, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "398.53", "answer_float": 398.53, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily dry matter increment (日干物质增量).\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "398.53", "answer_float": 398.53, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "398.53", "answer_float": 398.53, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "398.53", "answer_float": 398.53, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Physiological age factor (生理年龄因子)\nFormula (LaTeX): F_{PA}=1\\;(PDT<28),\\quad e^{-a(PDT-28)}\\;(PDT\\ge 28)\nFormula: FPA = if PDT<28: 1; else: exp(-a*(PDT-28))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[12.0, 57.0]\n- a: senescence decline coefficient; unit=dimensionless; typical range=[0.015, 0.07]\nPhysiological category: Photosynthesis\nMathematical form: Piecewise exponential decay\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "398.53", "answer_float": 398.53, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF033", "formula_name_en": "Daily dry matter increment", "id": "CMF033_chain_2_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Subtraction and conversion", "mode": "chain_2", "precision": 2, "problem": "In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Actual maximum assimilation rate (实际最大同化速率)\nFormula (LaTeX): AMAX=Am\\cdot F_{CO2}\\cdot F_{PA}\\cdot F_T\\cdot\\min(NNI,WDF)\nFormula: AMAX = Am * FCO2 * FPA * FT * min(NNI, WDF)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg CO2/ha/h\nParameters:\n- Am: genetic maximum assimilation rate; unit=kg CO2/ha/h; typical range=[25.0, 55.0]\n- FCO2: CO2 factor; unit=fraction; typical range=[0.95, 1.35]\n- FPA: physiological age factor; unit=fraction; typical range=[0.45, 1.0]\n- FT: temperature factor; unit=fraction; typical range=[0.45, 1.0]\n- NNI: nitrogen nutrition index; unit=fraction; typical range=[0.45, 1.1]\n- WDF: water deficit factor; unit=fraction; typical range=[0.45, 1.1]\nPhysiological category: Photosynthesis\nMathematical form: Multiplicative with minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In respiration, the Daily dry matter increment is obtained by chaining intermediate calculations. With inputs DTGA=502.1 kg/ha/day, RM=62.3 kg/ha/day, RG=61.2 kg/ha/day, b=0.05 fraction, determine the numeric value in kg/ha/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF033_chain_2_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "kg/ha/day"} +{"answer": "0.792", "answer_float": 0.792, "change": {"new_value": 0.05, "old_value": 0.045, "parameter": "Ratio_N_C_photo_product"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_sensitivity_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.792", "answer_float": 0.792, "change": {"new_value": 0.05, "old_value": 0.045, "parameter": "Ratio_N_C_photo_product"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_sensitivity_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.792", "answer_float": 0.792, "change": {"new_value": 0.05, "old_value": 0.045, "parameter": "Ratio_N_C_photo_product"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.792", "answer_float": 0.792, "change": {"new_value": 0.05, "old_value": 0.045, "parameter": "Ratio_N_C_photo_product"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.792", "answer_float": 0.792, "change": {"new_value": 0.05, "old_value": 0.045, "parameter": "Ratio_N_C_photo_product"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling canopy absorbed PAR (冠层吸收光合有效辐射)\nFormula (LaTeX): PAR=Radi\\cdot0.44\\cdot4.6(1-\\rho)\\frac{1-e^{-KLAI}}{KLAI}\nFormula: PAR = Radi * 0.44 * 4.6 * (1-Coef_rou) * (1-exp(-Coef_K*LAI)) / (Coef_K*LAI)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol/m2/s\nParameters:\n- Radi: incoming radiation; unit=W/m2; typical range=[120.0, 720.0]\n- LAI: leaf area index; unit=m2/m2; typical range=[0.4, 6.5]\n- Coef_rou: canopy reflectance; unit=fraction; typical range=[0.08, 0.25]\n- Coef_K: extinction coefficient; unit=dimensionless; typical range=[0.35, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Exponential ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.792", "answer_float": 0.792, "change": {"new_value": 0.05, "old_value": 0.045, "parameter": "Ratio_N_C_photo_product"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise minimum", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements N_shoot=5.24 g N/m2, Coef_Plant_TN_uptake=0.117 d-1, N_requirement_max_day=2.09 g N/m2/day, NPP=15.84 g C/m2/day, Ratio_N_C_photo_product=0.045 g N/g C, DVI=0.49 dimensionless in carbon-nitrogen cycling nitrogen, what would the carbon-nitrogen cycling crop nitrogen demand be if Ratio_N_C_photo_product were adjusted to 0.05? Provide the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 1, "unit": "g N/m2/day"} +{"answer": "0.481", "answer_float": 0.481, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.481", "answer_float": 0.481, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.481", "answer_float": 0.481, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.481", "answer_float": 0.481, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.481", "answer_float": 0.481, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Panicle partitioning index (穗分配指数)\nFormula (LaTeX): PIP=\\frac{HI/0.87}{1+e^{-0.2804(PDT-39)}}\nFormula: PIP = (Y_grain / B_above / 0.87) / (1 + exp(-0.2804 * (PDT - 39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Y_grain: grain yield; unit=kg/ha; typical range=[3500.0, 9000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[9000.0, 19000.0]\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Multi-step sigmoid\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "0.481", "answer_float": 0.481, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_direct_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Stage-dependent product", "mode": "direct", "precision": 3, "problem": "Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Maintenance respiration (维持呼吸)\nFormula (LaTeX): RM=(0.0091-0.0001\\cdot PDT)\\cdot ABIOMASS\\cdot Q_{10}^{(T-T_o)/10}\nFormula: RM = (0.0091 - 0.0001 * PDT) * ABIOMASS * Q10**((T-To)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n- ABIOMASS: above-ground biomass; unit=kg/ha; typical range=[2000.0, 18000.0]\n- Q10: respiration temperature coefficient; unit=dimensionless; typical range=[1.7, 2.3]\n- T: current temperature; unit=C; typical range=[16.0, 38.0]\n- To: optimum temperature; unit=C; typical range=[28.0, 30.0]\nPhysiological category: Respiration\nMathematical form: Product with Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given PDT=39.5 physiological days, DTE=0.668 fraction, IE=0.67 dimensionless, RPE=0.5 fraction, BFF=0.72 dimensionless, calculate the Daily physiological effectiveness for a phenology. The result should be in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 1, "unit": "physiological days/day"} +{"answer": "11.269", "answer_float": 11.269, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "11.269", "answer_float": 11.269, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "11.269", "answer_float": 11.269, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "11.269", "answer_float": 11.269, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "11.269", "answer_float": 11.269, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "11.269", "answer_float": 11.269, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Maximum with chained drivers", "mode": "direct", "precision": 3, "problem": "In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Methane production is computed as part of methane modeling. With FEh=0.601 fraction, TI=0.715 fraction, Cr=88.53 g C/m2/day, Com=6.15 g C/m2/day, determine the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.0084", "answer_float": 0.0084, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0001__C", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0084", "answer_float": 0.0084, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0001__K_name", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数).\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0084", "answer_float": 0.0084, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0084", "answer_float": 0.0084, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0084", "answer_float": 0.0084, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Relative thermal effectiveness (相对热效应)\nFormula (LaTeX): RTE=\\left[\\frac{T-T_b}{T_o-T_b}\\left(\\frac{T_m-T}{T_m-T_o}\\right)^{(T_m-T_o)/(T_o-T_b)}\\right]^{TS}\nFormula: RTE = (((T - Tb)/(To - Tb)) * (((Tm - T)/(Tm - To)) ** ((Tm - To)/(To - Tb)))) ** TS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- T: hourly temperature; unit=C; typical range=[12.0, 40.0]\n- Tb: base temperature; unit=C; typical range=[10.0, 10.0]\n- To: optimum temperature; unit=C; typical range=[30.0, 30.0]\n- Tm: maximum temperature; unit=C; typical range=[42.0, 42.0]\n- TS: temperature sensitivity; unit=dimensionless; typical range=[0.8, 1.8]\nPhysiological category: Phenology\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Maintenance respiration coefficient at optimum temperature (最适温度维持呼吸系数)\nFormula (LaTeX): RM(T_o)=0.0091-0.0001PDT\nFormula: RMTo = 0.0091 - 0.0001 * PDT\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: d-1\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Respiration\nMathematical form: Linear decline\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.0084", "answer_float": 0.0084, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF049", "formula_name_en": "Maintenance respiration coefficient at optimum temperature", "id": "CMF049_direct_t2_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Linear decline", "mode": "direct", "precision": 4, "problem": "In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Daily dry matter increment (日干物质增量)\nFormula (LaTeX): GCR=\\frac{DTGA-RM-RG}{1-b}\nFormula: GCR = (DTGA - RM - RG) / (1 - b)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- DTGA: daily total gross assimilation; unit=kg/ha/day; typical range=[180.0, 620.0]\n- RM: maintenance respiration; unit=kg/ha/day; typical range=[20.0, 120.0]\n- RG: growth respiration; unit=kg/ha/day; typical range=[20.0, 160.0]\n- b: mineral fraction in dry matter conversion; unit=fraction; typical range=[0.05, 0.05]\nPhysiological category: Respiration\nMathematical form: Subtraction and conversion\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a respiration, the Maintenance respiration coefficient at optimum temperature is computed as part of respiration modeling. With PDT=6.8 physiological days, determine the numeric value in d-1.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF049_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Respiration", "template_id": 2, "unit": "d-1"} +{"answer": "0.680", "answer_float": 0.68, "change": null, "condition": "C", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0000__C", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.680", "answer_float": 0.68, "change": null, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0000__K_name", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Soil organic matter carbon decomposition (土壤有机质碳分解量).\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.680", "answer_float": 0.68, "change": null, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.680", "answer_float": 0.68, "change": null, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.680", "answer_float": 0.68, "change": null, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "0.680", "answer_float": 0.68, "change": null, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF021", "formula_name_en": "Soil organic matter carbon decomposition", "id": "CMF021_direct_t2_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Multiplicative", "mode": "direct", "precision": 3, "problem": "In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a methane, the Soil organic matter carbon decomposition is computed as part of methane modeling. With WI=0.486 fraction, SI=1.0 fraction, TI=0.56 fraction, OMS=832.8 g C/m2, determine the numeric value in g C/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF021_direct_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g C/m2/day"} +{"answer": "2.690", "answer_float": 2.69, "change": {"new_value": 1.315, "old_value": 1.461, "parameter": "TI"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_sensitivity_t1_0001__C", "is_gold": false, "level": 2, "math_type": "Maximum with chained drivers", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "2.690", "answer_float": 2.69, "change": {"new_value": 1.315, "old_value": 1.461, "parameter": "TI"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_sensitivity_t1_0001__K_name", "is_gold": false, "level": 2, "math_type": "Maximum with chained drivers", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Methane production (甲烷产生量).\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "2.690", "answer_float": 2.69, "change": {"new_value": 1.315, "old_value": 1.461, "parameter": "TI"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_sensitivity_t1_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Maximum with chained drivers", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "2.690", "answer_float": 2.69, "change": {"new_value": 1.315, "old_value": 1.461, "parameter": "TI"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_sensitivity_t1_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Maximum with chained drivers", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "2.690", "answer_float": 2.69, "change": {"new_value": 1.315, "old_value": 1.461, "parameter": "TI"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_sensitivity_t1_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Maximum with chained drivers", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Fresh organic matter carbon decomposition (新鲜有机质碳分解量)\nFormula (LaTeX): OMNC=WI\\cdot SI\\cdot TI\\cdot0.027\\cdot OMN\nFormula: OMNC = WI * SI * TI * 0.027 * OMN\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMN: fresh organic matter carbon pool; unit=g C/m2; typical range=[20.0, 260.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane production (甲烷产生量)\nFormula (LaTeX): P=\\max(0,0.27F_{Eh}(TI\\cdot C_r+Com))\nFormula: P = max(0, 0.27 * FEh * (TI * Cr + Com))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- FEh: redox factor; unit=fraction; typical range=[0.05, 1.0]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- Cr: root-exudate carbon substrate; unit=g C/m2/day; typical range=[1.0, 90.0]\n- Com: organic matter decomposition carbon; unit=g C/m2/day; typical range=[0.05, 8.0]\nPhysiological category: Methane\nMathematical form: Maximum with chained drivers\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "2.690", "answer_float": 2.69, "change": {"new_value": 1.315, "old_value": 1.461, "parameter": "TI"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF022", "formula_name_en": "Methane production", "id": "CMF022_sensitivity_t1_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Maximum with chained drivers", "mode": "sensitivity", "precision": 3, "problem": "Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane ebullition emission (甲烷冒泡排放)\nFormula (LaTeX): E_{bl}=\\min(0.7\\ln(t_{soil})/W_r,0.9)P\nFormula: Ebl = min(0.7 * ln(t_soil) / Wr, 0.9) * P\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- t_soil: soil temperature; unit=C; typical range=[12.0, 36.0]\n- Wr: root biomass; unit=kg/ha; typical range=[150.0, 2800.0]\nPhysiological category: Methane\nMathematical form: Logarithmic minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements FEh=0.702 fraction, TI=1.461 fraction, Cr=7.84 g C/m2/day, Com=3.883 g C/m2/day in methane, what would the Methane production be if TI were adjusted to 1.315? Provide the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF022_sensitivity_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 1, "unit": "g CH4/m2/day"} +{"answer": "0.091", "answer_float": 0.091, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t2_0001__C", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 2, "unit": "g N/m2/day"} +{"answer": "0.091", "answer_float": 0.091, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t2_0001__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量).\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 2, "unit": "g N/m2/day"} +{"answer": "0.091", "answer_float": 0.091, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t2_0001__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 2, "unit": "g N/m2/day"} +{"answer": "0.091", "answer_float": 0.091, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t2_0001__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 2, "unit": "g N/m2/day"} +{"answer": "0.091", "answer_float": 0.091, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t2_0001__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop nitrogen demand (作物需氮量)\nFormula (LaTeX): N_d=C_uN_{shoot}\\;\\text{or}\\;\\min(N_{max},r_NNPP)\nFormula: N_demand = if NPP<0: 0; if N_shoot<3.5 and DVI<0.3: Coef_Plant_TN_uptake*N_shoot; else: min(N_requirement_max_day, Ratio_N_C_photo_product*NPP)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2/day\nParameters:\n- N_shoot: shoot nitrogen; unit=g N/m2; typical range=[0.8, 8.0]\n- Coef_Plant_TN_uptake: plant nitrogen uptake coefficient; unit=d-1; typical range=[0.04, 0.22]\n- N_requirement_max_day: daily maximum nitrogen requirement; unit=g N/m2/day; typical range=[0.4, 3.0]\n- NPP: net primary production; unit=g C/m2/day; typical range=[1.0, 18.0]\n- Ratio_N_C_photo_product: nitrogen-carbon ratio of photosynthate; unit=g N/g C; typical range=[0.01, 0.06]\n- DVI: development index; unit=dimensionless; typical range=[0.0, 1.8]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Piecewise minimum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 2, "unit": "g N/m2/day"} +{"answer": "0.091", "answer_float": 0.091, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF004", "formula_name_en": "carbon-nitrogen cycling crop nitrogen demand", "id": "CMF004_direct_t2_0001__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise minimum", "mode": "direct", "precision": 3, "problem": "In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In a carbon-nitrogen cycling nitrogen, the carbon-nitrogen cycling crop nitrogen demand is computed as part of carbon-nitrogen cycling nitrogen modeling. With N_shoot=6.78 g N/m2, Coef_Plant_TN_uptake=0.166 d-1, N_requirement_max_day=1.64 g N/m2/day, NPP=1.55 g C/m2/day, Ratio_N_C_photo_product=0.059 g N/g C, DVI=0.73 dimensionless, determine the numeric value in g N/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF004_direct_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling nitrogen", "template_id": 2, "unit": "g N/m2/day"} +{"answer": "1.392", "answer_float": 1.392, "change": null, "condition": "C", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_branch_t1_0000__C", "is_gold": false, "level": 4, "math_type": "Piecewise linear cap", "mode": "branch", "precision": 3, "problem": "For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.392", "answer_float": 1.392, "change": null, "condition": "K_name", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_branch_t1_0000__K_name", "is_gold": false, "level": 4, "math_type": "Piecewise linear cap", "mode": "branch", "precision": 3, "problem": "For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: environmental-response CO2 response (环境响应二氧化碳响应).\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.392", "answer_float": 1.392, "change": null, "condition": "K_formula", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_branch_t1_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Piecewise linear cap", "mode": "branch", "precision": 3, "problem": "For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.392", "answer_float": 1.392, "change": null, "condition": "K_domain", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_branch_t1_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Piecewise linear cap", "mode": "branch", "precision": 3, "problem": "For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.392", "answer_float": 1.392, "change": null, "condition": "K_distractor", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_branch_t1_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Piecewise linear cap", "mode": "branch", "precision": 3, "problem": "For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: environmental-response CO2 response (环境响应二氧化碳响应)\nFormula (LaTeX): F_{CO2}=f(CO2,CO2_{RUE})\nFormula: F_CO2 = if CO2>=700: 1+CO2_RUE*350/100; else: max(CO2_RUE*CO2*0.01+1-0.01*350*CO2_RUE, 1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: carbon dioxide concentration; unit=ppm; typical range=[330.0, 820.0]\n- CO2_RUE: CO2 response coefficient for RUE; unit=fraction; typical range=[0.05, 0.18]\nPhysiological category: environmental-response formula system\nMathematical form: Piecewise linear cap\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: environmental-response daily biomass increment (环境响应日生物量增量)\nFormula (LaTeX): dB=10RUEfSolarSRADF_{CO2}F_T\\min(F_W,F_H)\nFormula: dBiomass = 10 * RUE * fSolar * SRAD * F_CO2 * F_Temp * min(F_Water, F_Heat)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha/day\nParameters:\n- RUE: radiation use efficiency; unit=g/MJ; typical range=[1.1, 3.8]\n- fSolar: intercepted solar fraction; unit=fraction; typical range=[0.15, 0.95]\n- SRAD: solar radiation; unit=MJ/m2/day; typical range=[8.0, 28.0]\n- F_CO2: CO2 response factor; unit=fraction; typical range=[1.0, 1.35]\n- F_Temp: temperature response factor; unit=fraction; typical range=[0.35, 1.0]\n- F_Water: water response factor; unit=fraction; typical range=[0.25, 1.0]\n- F_Heat: heat response factor; unit=fraction; typical range=[0.25, 1.0]\nPhysiological category: environmental-response formula system\nMathematical form: Multiplicative minimum stress\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "1.392", "answer_float": 1.392, "change": null, "condition": "K_wrong", "family": "environmental_response", "formula_id": "CMF057", "formula_name_en": "environmental-response CO2 response", "id": "CMF057_branch_t1_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Piecewise linear cap", "mode": "branch", "precision": 3, "problem": "For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: environmental-response daily thermal time (每日有效积温)\nFormula (LaTeX): dTT=\\max(T_{mean}-T_{base},0)\nFormula: dTT = max(Tmean - Tbase, 0)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: C day\nParameters:\n- Tmean: mean temperature; unit=C; typical range=[0.0, 34.0]\n- Tbase: base temperature; unit=C; typical range=[4.0, 12.0]\nPhysiological category: environmental-response formula system\nMathematical form: Maximum linear\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: For environmental-response formula system, compute the environmental-response CO2 response. The inputs are CO2=703.0 ppm, CO2_RUE=0.112 fraction. Note that this formula uses piecewise logic depending on parameter values. Report the final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF057_branch_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "environmental-response formula system", "template_id": 1, "unit": "multiplier"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 40.7, "old_value": 37.0, "parameter": "T"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0000__C", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 40.7, "old_value": 37.0, "parameter": "T"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0000__K_name", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Temperature response factor (温度响应因子).\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 40.7, "old_value": 37.0, "parameter": "T"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 40.7, "old_value": 37.0, "parameter": "T"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 40.7, "old_value": 37.0, "parameter": "T"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Root partitioning index (地下部分配指数)\nFormula (LaTeX): PIRO=1-PISH\nFormula: PIRO = 1 - (-0.0000842*PDT^2 + 0.01*PDT + 0.63)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual from shoot partitioning\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "0.58", "answer_float": 0.58, "change": {"new_value": 40.7, "old_value": 37.0, "parameter": "T"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF032", "formula_name_en": "Temperature response factor", "id": "CMF032_sensitivity_t1_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Piecewise trigonometric", "mode": "sensitivity", "precision": 2, "problem": "Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given current measurements T=37.0 C, Tb=10.0 C, Tol=24.0 C, Tou=34.0 C, Tmax=45.0 C in temperature response, what would the Temperature response factor be if T were adjusted to 40.7? Provide the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF032_sensitivity_t1_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Temperature response", "template_id": 1, "unit": "fraction"} +{"answer": "90.97", "answer_float": 90.97, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t0_0000__C", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "90.97", "answer_float": 90.97, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t0_0000__K_name", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "90.97", "answer_float": 90.97, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t0_0000__K_formula", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "90.97", "answer_float": 90.97, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t0_0000__K_domain", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "90.97", "answer_float": 90.97, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t0_0000__K_distractor", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解���子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "90.97", "answer_float": 90.97, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_direct_t0_0000__K_wrong", "is_gold": false, "level": 1, "math_type": "Ratio", "mode": "direct", "precision": 2, "problem": "A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation estimates Nitrogen use efficiency. The measured inputs are GrainYield=8269.0 kg/ha, NUptake=90.9 kg N/ha. What is the nitrogen use efficiency? Give the final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_direct_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "22.164", "answer_float": 22.164, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t0_0001__C", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol CO2/m2/s"} +{"answer": "22.164", "answer_float": 22.164, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t0_0001__K_name", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率).\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol CO2/m2/s"} +{"answer": "22.164", "answer_float": 22.164, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t0_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol CO2/m2/s"} +{"answer": "22.164", "answer_float": 22.164, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t0_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol CO2/m2/s"} +{"answer": "22.164", "answer_float": 22.164, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t0_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling optimal photosynthetic rate (最适光合速率)\nFormula (LaTeX): P_{opt}=\\alpha N_{leaf}\\frac{PAR}{PAR+\\beta}\nFormula: Photo_Rate_opt = (Coef_alfa * Leaf_N) * PAR / (PAR + Coef_beta)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: umol CO2/m2/s\nParameters:\n- Coef_alfa: maximum photosynthesis coefficient; unit=umol CO2/gN/s; typical range=[8.0, 28.0]\n- Coef_beta: half-saturation radiation; unit=umol/m2/s; typical range=[80.0, 360.0]\n- PAR: absorbed PAR; unit=umol/m2/s; typical range=[80.0, 1200.0]\n- Leaf_N: leaf nitrogen content; unit=gN/m2; typical range=[0.6, 4.5]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Michaelis-Menten response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol CO2/m2/s"} +{"answer": "22.164", "answer_float": 22.164, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF007", "formula_name_en": "carbon-nitrogen cycling optimal photosynthetic rate", "id": "CMF007_direct_t0_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Michaelis-Menten response", "mode": "direct", "precision": 3, "problem": "A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A carbon-nitrogen cycling photosynthesis calculation estimates carbon-nitrogen cycling optimal photosynthetic rate. The measured inputs are Coef_alfa=19.43 umol CO2/gN/s, Coef_beta=354.5 umol/m2/s, PAR=920.5 umol/m2/s, Leaf_N=1.58 gN/m2. What is the carbon-nitrogen cycling optimal photosynthetic rate? Give the final numeric value in umol CO2/m2/s.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF007_direct_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 0, "unit": "umol CO2/m2/s"} +{"answer": "1.45", "answer_float": 1.45, "change": {"new_value": 0.81, "old_value": 0.9, "parameter": "beta"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0001__C", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.45", "answer_float": 1.45, "change": {"new_value": 0.81, "old_value": 0.9, "parameter": "beta"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0001__K_name", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: CO2 response factor (二氧化碳影响因子).\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.45", "answer_float": 1.45, "change": {"new_value": 0.81, "old_value": 0.9, "parameter": "beta"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.45", "answer_float": 1.45, "change": {"new_value": 0.81, "old_value": 0.9, "parameter": "beta"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.45", "answer_float": 1.45, "change": {"new_value": 0.81, "old_value": 0.9, "parameter": "beta"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: CO2 response factor (二氧化碳影响因子)\nFormula (LaTeX): F_{CO2}=1+\\beta\\ln(C_x/340)\nFormula: FCO2 = 1 + beta * ln(Cx / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Cx: atmospheric CO2 concentration; unit=ppm; typical range=[360.0, 720.0]\n- beta: CO2 response coefficient; unit=dimensionless; typical range=[0.35, 0.95]\nPhysiological category: Photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Green leaf partitioning index (绿叶分配指数)\nFormula (LaTeX): PIGL=0.54-0.0046PDT\\;(PDT<26),\\quad1.4532e^{-0.0492PDT}\\;(PDT\\ge26)\nFormula: PIGL = if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Piecewise linear/exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "1.45", "answer_float": 1.45, "change": {"new_value": 0.81, "old_value": 0.9, "parameter": "beta"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF029", "formula_name_en": "CO2 response factor", "id": "CMF029_sensitivity_t0_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Logarithmic", "mode": "sensitivity", "precision": 2, "problem": "A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Dry grain yield (稻谷干重产量)\nFormula (LaTeX): Y=AWSP\\cdot f_{grain}\nFormula: Y = AWSP * grain_fraction\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg/ha\nParameters:\n- AWSP: actual panicle dry weight; unit=kg/ha; typical range=[3500.0, 10500.0]\n- grain_fraction: dry grain fraction of panicle weight; unit=fraction; typical range=[0.82, 0.92]\nPhysiological category: Yield formation\nMathematical form: Product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A photosynthesis calculation involves sensitivity analysis. The current measurements are Cx=593.0 ppm, beta=0.9 dimensionless. A field technician asks what the co2 response factor would become if beta changed to 0.81. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF029_sensitivity_t0_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Photosynthesis", "template_id": 0, "unit": "multiplier"} +{"answer": "8.457", "answer_float": 8.457, "change": {"new_value": 15.106, "old_value": 16.784, "parameter": "P"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_sensitivity_t2_0000__C", "is_gold": false, "level": 3, "math_type": "Chained sum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.457", "answer_float": 8.457, "change": {"new_value": 15.106, "old_value": 16.784, "parameter": "P"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_sensitivity_t2_0000__K_name", "is_gold": false, "level": 3, "math_type": "Chained sum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Total methane emission (甲烷总排放).\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.457", "answer_float": 8.457, "change": {"new_value": 15.106, "old_value": 16.784, "parameter": "P"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 3, "math_type": "Chained sum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.457", "answer_float": 8.457, "change": {"new_value": 15.106, "old_value": 16.784, "parameter": "P"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 3, "math_type": "Chained sum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.457", "answer_float": 8.457, "change": {"new_value": 15.106, "old_value": 16.784, "parameter": "P"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 3, "math_type": "Chained sum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Total methane emission (甲烷总排放)\nFormula (LaTeX): E=E_{bl}+P\\cdot C(1-W/W_{max})^{0.25}\nFormula: E = Ebl + P * CH4RiceEfC * (1 - W/Wmax)^0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g CH4/m2/day\nParameters:\n- Ebl: ebullition emission; unit=g CH4/m2/day; typical range=[0.01, 6.0]\n- P: methane production; unit=g CH4/m2/day; typical range=[0.5, 18.0]\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Chained sum\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "8.457", "answer_float": 8.457, "change": {"new_value": 15.106, "old_value": 16.784, "parameter": "P"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF017", "formula_name_en": "Total methane emission", "id": "CMF017_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 3, "math_type": "Chained sum", "mode": "sensitivity", "precision": 3, "problem": "In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Methane temperature index (甲烷温度指数)\nFormula (LaTeX): TI=Q_{10}^{(t_{sl}-30)/10}\nFormula: TI = Q10^((min(t_soil,30)-30)/10)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Q10: temperature coefficient; unit=dimensionless; typical range=[2.0, 4.0]\n- t_soil: soil temperature; unit=C; typical range=[15.0, 42.0]\nPhysiological category: Methane\nMathematical form: Q10 with temperature clipping\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In methane, we analyze how Total methane emission responds to parameter changes. Baseline: Ebl=2.514 g CH4/m2/day, P=16.784 g CH4/m2/day, CH4RiceEfC=0.468 fraction, W=6006.3 kg/ha, Wmax=11997.5 kg/ha. After changing P from 16.784 to 15.106, compute the new Total methane emission. Give the numeric value in g CH4/m2/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF017_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 2, "unit": "g CH4/m2/day"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "C", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0000__C", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0000__K_name", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Daily physiological effectiveness (每日生理效应).\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0000__K_formula", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0000__K_domain", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0000__K_distractor", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Daily physiological effectiveness (每日生理效应)\nFormula (LaTeX): DPE=DTE\\cdot IE,\\;DTE\\cdot RPE,\\;DTE,\\;DTE\\cdot BFF\nFormula: DPE = if PDT<8: DTE*IE; if PDT<18: DTE*RPE; if PDT<32: DTE; if PDT<57: DTE*BFF; else: 0\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: physiological days/day\nParameters:\n- PDT: current physiological development time; unit=physiological days; typical range=[0.0, 56.0]\n- DTE: daily thermal effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- IE: intrinsic earliness factor; unit=dimensionless; typical range=[0.65, 1.25]\n- RPE: relative photoperiod effectiveness; unit=fraction; typical range=[0.25, 1.0]\n- BFF: basic filling factor; unit=dimensionless; typical range=[0.65, 1.25]\nPhysiological category: Phenology\nMathematical form: Stage-dependent product\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Temperature response factor (温度响应因子)\nFormula (LaTeX): F_T=\\sin(\\frac{T-T_b}{T_{ol}-T_b}\\frac{\\pi}{2}); 1; \\cos(\\frac{T-T_{ou}}{T_{max}-T_{ou}}\\frac{\\pi}{2})\nFormula: F_T = if Tb<=T\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "0.241", "answer_float": 0.241, "change": null, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF027", "formula_name_en": "Daily physiological effectiveness", "id": "CMF027_branch_t0_0000__K_wrong", "is_gold": false, "level": 4, "math_type": "Stage-dependent product", "mode": "branch", "precision": 3, "problem": "A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Low-temperature fertility factor (低温结实因子)\nFormula (LaTeX): LTF=1-\\frac{4.6+0.054Q_t^{1.56}}{100}\nFormula: LTF = if 26<=PDT<=39 and Tmean<22: 1-(4.6+0.054*Qt^1.56)/100; else: 1\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[26.0, 39.0]\n- Tmean: daily mean temperature; unit=C; typical range=[16.0, 23.0]\n- Qt: accumulated low-temperature degree-days; unit=C day; typical range=[1.0, 24.0]\nPhysiological category: Temperature stress\nMathematical form: Piecewise power penalty\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A phenology involves a stage that may fall on either side of a piecewise threshold. The measured inputs are PDT=53.3 physiological days, DTE=0.274 fraction, IE=0.91 dimensionless, RPE=0.94 fraction, BFF=0.88 dimensionless. Determine the daily physiological effectiveness and report the final numeric value in physiological days/day.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF027_branch_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Phenology", "template_id": 0, "unit": "physiological days/day"} +{"answer": "39.02", "answer_float": 39.02, "change": {"new_value": 100.4, "old_value": 111.5, "parameter": "NUptake"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "39.02", "answer_float": 39.02, "change": {"new_value": 100.4, "old_value": 111.5, "parameter": "NUptake"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "39.02", "answer_float": 39.02, "change": {"new_value": 100.4, "old_value": 111.5, "parameter": "NUptake"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "39.02", "answer_float": 39.02, "change": {"new_value": 100.4, "old_value": 111.5, "parameter": "NUptake"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "39.02", "answer_float": 39.02, "change": {"new_value": 100.4, "old_value": 111.5, "parameter": "NUptake"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "39.02", "answer_float": 39.02, "change": {"new_value": 100.4, "old_value": 111.5, "parameter": "NUptake"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling shoot nitrogen allocation (地上部氮分配量)\nFormula (LaTeX): N_{shoot}=N_{uptake}\\frac{B_{shoot}}{B_{shoot}+B_{root}}\nFormula: N_shoot = Plant_TN_uptake * Biom_shoot / (Biom_shoot + Biom_root)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g N/m2\nParameters:\n- Plant_TN_uptake: total plant nitrogen uptake; unit=g N/m2; typical range=[1.0, 18.0]\n- Biom_shoot: shoot biomass; unit=g C/m2; typical range=[80.0, 850.0]\n- Biom_root: root biomass; unit=g C/m2; typical range=[20.0, 320.0]\nPhysiological category: carbon-nitrogen cycling nitrogen\nMathematical form: Biomass-weighted allocation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A nitrogen calculation involves sensitivity analysis. The current measurements are GrainYield=3918.0 kg/ha, NUptake=111.5 kg N/ha. A field technician asks what the nitrogen use efficiency would become if NUptake changed to 100.4. Report the new final numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 0, "unit": "kg grain/kg N"} +{"answer": "0.71", "answer_float": 0.71, "change": {"new_value": 8.3, "old_value": 9.2, "parameter": "PDT"}, "condition": "C", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t2_0000__C", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.71", "answer_float": 0.71, "change": {"new_value": 8.3, "old_value": 9.2, "parameter": "PDT"}, "condition": "K_name", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t2_0000__K_name", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Shoot partitioning index (地上部分配指数).\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.71", "answer_float": 0.71, "change": {"new_value": 8.3, "old_value": 9.2, "parameter": "PDT"}, "condition": "K_formula", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t2_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.71", "answer_float": 0.71, "change": {"new_value": 8.3, "old_value": 9.2, "parameter": "PDT"}, "condition": "K_domain", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t2_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.71", "answer_float": 0.71, "change": {"new_value": 8.3, "old_value": 9.2, "parameter": "PDT"}, "condition": "K_distractor", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t2_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Potential leaf area index (潜在叶面积指数)\nFormula (LaTeX): LAI_P=LAI_0e^{RP\\cdot GDD}\nFormula: LAIP = LAI0 * exp(RP * GDD)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: m2/m2\nParameters:\n- LAI0: initial leaf area index; unit=m2/m2; typical range=[0.05, 0.35]\n- RP: potential relative growth rate for LAI; unit=d-1; typical range=[0.003, 0.008]\n- GDD: growing degree days; unit=C day; typical range=[120.0, 680.0]\nPhysiological category: Leaf area growth\nMathematical form: Exponential growth\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Shoot partitioning index (地上部分配指数)\nFormula (LaTeX): PISH=-8.42\\times10^{-5}PDT^2+0.01PDT+0.63\nFormula: PISH = -0.0000842 * PDT^2 + 0.01 * PDT + 0.63\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[0.0, 57.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Quadratic polynomial\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.71", "answer_float": 0.71, "change": {"new_value": 8.3, "old_value": 9.2, "parameter": "PDT"}, "condition": "K_wrong", "family": "growth_yield", "formula_id": "CMF044", "formula_name_en": "Shoot partitioning index", "id": "CMF044_sensitivity_t2_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Quadratic polynomial", "mode": "sensitivity", "precision": 2, "problem": "In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Stem partitioning index (茎鞘分配指数)\nFormula (LaTeX): PIS=1-PIGL-PIP\nFormula: PIS = 1 - (if PDT<26: 0.54-0.0046*PDT; else: 1.4532*exp(-0.0492*PDT)) - (Y_grain/B_above/0.87)/(1+exp(-0.2804*(PDT-39)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- PDT: physiological development time; unit=physiological days; typical range=[24.0, 57.0]\n- Y_grain: grain yield used to derive potential panicle partitioning; unit=kg/ha; typical range=[3500.0, 8000.0]\n- B_above: above-ground biomass; unit=kg/ha; typical range=[10500.0, 19000.0]\nPhysiological category: Dry matter partitioning\nMathematical form: Residual with piecewise components\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In dry matter partitioning, we analyze how Shoot partitioning index responds to parameter changes. Baseline: PDT=9.2 physiological days. After changing PDT from 9.2 to 8.3, compute the new Shoot partitioning index. Give the numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF044_sensitivity_t2_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Dry matter partitioning", "template_id": 2, "unit": "fraction"} +{"answer": "0.853", "answer_float": 0.853, "change": {"new_value": 0.522, "old_value": 0.58, "parameter": "WaterC"}, "condition": "C", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t0_0000__C", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.853", "answer_float": 0.853, "change": {"new_value": 0.522, "old_value": 0.58, "parameter": "WaterC"}, "condition": "K_name", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t0_0000__K_name", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Water-content decomposition factor (水分分解因子).\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.853", "answer_float": 0.853, "change": {"new_value": 0.522, "old_value": 0.58, "parameter": "WaterC"}, "condition": "K_formula", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t0_0000__K_formula", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.853", "answer_float": 0.853, "change": {"new_value": 0.522, "old_value": 0.58, "parameter": "WaterC"}, "condition": "K_domain", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t0_0000__K_domain", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.853", "answer_float": 0.853, "change": {"new_value": 0.522, "old_value": 0.58, "parameter": "WaterC"}, "condition": "K_distractor", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t0_0000__K_distractor", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Water-content decomposition factor (水分分解因子)\nFormula (LaTeX): WI=0.49e^{3.88W_c-5.4W_c^2}\nFormula: WI = 0.49 * exp(3.88 * WaterC - 5.4 * WaterC^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- WaterC: soil water content index; unit=fraction; typical range=[0.25, 0.75]\nPhysiological category: Methane\nMathematical form: Exponential quadratic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: Plant methane transport factor (植株传输因子)\nFormula (LaTeX): F_p=C(1-W/W_{max})^{0.25}\nFormula: Fp = CH4RiceEfC * (1 - W / Wmax) ** 0.25\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CH4RiceEfC: plant transport coefficient; unit=fraction; typical range=[0.15, 0.55]\n- W: current shoot biomass; unit=kg/ha; typical range=[500.0, 7000.0]\n- Wmax: maximum shoot biomass; unit=kg/ha; typical range=[8000.0, 14000.0]\nPhysiological category: Methane\nMathematical form: Power attenuation\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "0.853", "answer_float": 0.853, "change": {"new_value": 0.522, "old_value": 0.58, "parameter": "WaterC"}, "condition": "K_wrong", "family": "methane_emission", "formula_id": "CMF025", "formula_name_en": "Water-content decomposition factor", "id": "CMF025_sensitivity_t0_0000__K_wrong", "is_gold": false, "level": 2, "math_type": "Exponential quadratic", "mode": "sensitivity", "precision": 3, "problem": "A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: Soil organic matter carbon decomposition (土壤有机质碳分解量)\nFormula (LaTeX): OMSC=WI\\cdot SI\\cdot TI\\cdot0.003\\cdot OMS\nFormula: OMSC = WI * SI * TI * 0.003 * OMS\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: g C/m2/day\nParameters:\n- WI: water-content factor; unit=fraction; typical range=[0.25, 0.7]\n- SI: soil index; unit=fraction; typical range=[0.6, 1.4]\n- TI: temperature index; unit=fraction; typical range=[0.25, 1.5]\n- OMS: soil organic matter carbon pool; unit=g C/m2; typical range=[250.0, 1800.0]\nPhysiological category: Methane\nMathematical form: Multiplicative\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: A methane calculation involves sensitivity analysis. The current measurements are WaterC=0.58 fraction. A field technician asks what the water-content decomposition factor would become if WaterC changed to 0.522. Report the new final numeric value in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF025_sensitivity_t0_0000", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Methane", "template_id": 0, "unit": "fraction"} +{"answer": "21.82", "answer_float": 21.82, "change": {"new_value": 3500.0, "old_value": 3725.0, "parameter": "GrainYield"}, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t2_0001__C", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 2, "unit": "kg grain/kg N"} +{"answer": "21.82", "answer_float": 21.82, "change": {"new_value": 3500.0, "old_value": 3725.0, "parameter": "GrainYield"}, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t2_0001__K_name", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nRelevant formula name: Nitrogen use efficiency (氮利用效率).\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 2, "unit": "kg grain/kg N"} +{"answer": "21.82", "answer_float": 21.82, "change": {"new_value": 3500.0, "old_value": 3725.0, "parameter": "GrainYield"}, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t2_0001__K_formula", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 2, "unit": "kg grain/kg N"} +{"answer": "21.82", "answer_float": 21.82, "change": {"new_value": 3500.0, "old_value": 3725.0, "parameter": "GrainYield"}, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t2_0001__K_domain", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 2, "unit": "kg grain/kg N"} +{"answer": "21.82", "answer_float": 21.82, "change": {"new_value": 3500.0, "old_value": 3725.0, "parameter": "GrainYield"}, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t2_0001__K_distractor", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: Nitrogen use efficiency (氮利用效率)\nFormula (LaTeX): NUE=\\frac{GrainYield}{NUptake}\nFormula: NUE = GrainYield / NUptake\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: kg grain/kg N\nParameters:\n- GrainYield: grain yield; unit=kg/ha; typical range=[3500.0, 9500.0]\n- NUptake: crop nitrogen uptake; unit=kg N/ha; typical range=[60.0, 210.0]\nPhysiological category: Nitrogen\nMathematical form: Ratio\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling soil carbon water factor (土壤碳分解水分因子)\nFormula (LaTeX): F_W=0.49e^{3.88W-5.4W^2}\nFormula: F_W_soil = if Soil_Water>=Saturate_Water: 0.65; else: 0.49*exp(3.88*(Soil_Water/BD)-5.4*(Soil_Water/BD)^2)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Soil_Water: soil water content; unit=cm3/cm3; typical range=[0.08, 0.55]\n- Saturate_Water: saturated water content; unit=cm3/cm3; typical range=[0.48, 0.48]\n- BD: bulk density; unit=g/cm3; typical range=[1.05, 1.55]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Piecewise exponential\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 2, "unit": "kg grain/kg N"} +{"answer": "21.82", "answer_float": 21.82, "change": {"new_value": 3500.0, "old_value": 3725.0, "parameter": "GrainYield"}, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF003", "formula_name_en": "Nitrogen use efficiency", "id": "CMF003_sensitivity_t2_0001__K_wrong", "is_gold": false, "level": 2, "math_type": "Ratio", "mode": "sensitivity", "precision": 2, "problem": "In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil carbon temperature factor (土壤碳分解温度因子)\nFormula (LaTeX): F_T=2.4^{(T_{soil}-10)/10}\\;\\text{where}\\;T_{soil}=T\\;\\text{or}\\;4.4+0.76T\nFormula: F_T_soil = if is_soil_temp>=0.5: 2.4**((Temp-10)*0.1); else: 2.4**((4.4+0.76*Temp-10)*0.1)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Temp: soil temperature or air temperature; unit=C; typical range=[5.0, 34.0]\n- is_soil_temp: 1 if Temp is soil temperature, 0 if Temp is air temperature; unit=0/1; typical range=[0.0, 1.0]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Q10 temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: In nitrogen, we analyze how Nitrogen use efficiency responds to parameter changes. Baseline: GrainYield=3725.0 kg/ha, NUptake=160.4 kg N/ha. After changing GrainYield from 3725.0 to 3500.0, compute the new Nitrogen use efficiency. Give the numeric value in kg grain/kg N.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF003_sensitivity_t2_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "Nitrogen", "template_id": 2, "unit": "kg grain/kg N"} +{"answer": "1.520", "answer_float": 1.52, "change": null, "condition": "C", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0001__C", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.520", "answer_float": 1.52, "change": null, "condition": "K_name", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0001__K_name", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include 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���子).\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.520", "answer_float": 1.52, "change": null, "condition": "K_formula", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0001__K_formula", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.520", "answer_float": 1.52, "change": null, "condition": "K_domain", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0001__K_domain", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.520", "answer_float": 1.52, "change": null, "condition": "K_distractor", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0001__K_distractor", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nCandidate formulas are listed below. Select the physiologically appropriate formula before calculating.\n\nCandidate 1:\nFormula name: carbon-nitrogen cycling CO2 fertilization factor (作物二氧化碳施肥因子)\nFormula (LaTeX): F_{CO2}=1+B\\ln(CO2/340)\nFormula: F_CO2 = 1 + Coef_B_CO2 * ln(CO2 / 340)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- CO2: atmospheric carbon dioxide concentration; unit=ppm; typical range=[330.0, 760.0]\n- Coef_B_CO2: CO2 response coefficient; unit=dimensionless; typical range=[0.2, 0.8]\nPhysiological category: carbon-nitrogen cycling photosynthesis\nMathematical form: Logarithmic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 2:\nFormula name: carbon-nitrogen cycling crop temperature activity factor (作物温度活性因子)\nFormula (LaTeX): F_T=\\left(\\frac{T_d-T_{min}}{T_{opt}-T_{min}}\\right)^{1+M}\\left(\\frac{T_{max}-T_d}{T_{max}-T_{opt}}\\right)^{1-M}\nFormula: F_T = if Td<=Tmin_active or Td>=Tmax_active: 0; else: ((Td-Tmin_active)/(Topt_active-Tmin_active))^(1+Coef_T_M) * ((Tmax_active-Td)/(Tmax_active-Topt_active))^(1-Coef_T_M)\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- Td: daytime crop temperature; unit=C; typical range=[6.0, 38.0]\n- Tmin_active: minimum active temperature; unit=C; typical range=[3.0, 3.0]\n- Topt_active: optimum active temperature; unit=C; typical range=[25.0, 25.0]\n- Tmax_active: maximum active temperature; unit=C; typical range=[42.0, 42.0]\n- Coef_T_M: temperature curve coefficient; unit=dimensionless; typical range=[0.1, 0.8]\nPhysiological category: carbon-nitrogen cycling stress\nMathematical form: Beta temperature response\nUse the formula only when the problem context matches the stated physiological quantity.\n\nCandidate 3:\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"} +{"answer": "1.520", "answer_float": 1.52, "change": null, "condition": "K_wrong", "family": "carbon_nitrogen_cycle", "formula_id": "CMF001", "formula_name_en": "carbon-nitrogen cycling CO2 fertilization factor", "id": "CMF001_direct_t1_0001__K_wrong", "is_gold": false, "level": 1, "math_type": "Logarithmic", "mode": "direct", "precision": 3, "problem": "Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.", "prompt": "<|im_start|>system\nYou are a scientific reasoning assistant. Solve the following problem step by step. Show your reasoning clearly, then end with exactly one line in this format: The answer is [number].\nDo not write multiple answers. Do not include units in the final line.\nExample:\nStep 1: ...\nStep 2: ...\nThe answer is 42.5<|im_end|>\n<|im_start|>user\nA retrieved formula is listed below. It may look relevant, but use it only if it truly matches the problem.\nFormula name: carbon-nitrogen cycling soil pH decomposition factor (土壤pH分解因子)\nFormula (LaTeX): F_{pH}=\\frac{1}{1+e^{2.5(5-pH)}}\nFormula: F_pH = 1 / (1 + exp(2.5 * (5 - pH)))\nOperator notes: * means multiplication, / means division, ** or ^ means power, exp(x) means e^x, ln(x) means natural logarithm, and min/max select the smaller/larger value.\nOutput unit: fraction\nParameters:\n- pH: soil pH; unit=pH; typical range=[3.8, 8.6]\nPhysiological category: carbon-nitrogen cycling soil carbon\nMathematical form: Logistic\nUse the formula only when the problem context matches the stated physiological quantity.\n\nQ: Given CO2=662.0 ppm, Coef_B_CO2=0.78 dimensionless, calculate the carbon-nitrogen cycling CO2 fertilization factor for a carbon-nitrogen cycling photosynthesis. The result should be in fraction.<|im_end|>\n<|im_start|>assistant\nA:", "sample_id": "CMF001_direct_t1_0001", "scoring_rule": "abs(prediction-answer_float) <= max(0.5*10^-precision, abs(answer_float)*0.01)", "split": "dev", "subject": "carbon-nitrogen cycling photosynthesis", "template_id": 1, "unit": "multiplier"}