[BUG] Got Zero em on MMLU using Deepseek distilled Qwen models #1047
Description
Describe the bug
Got Zero em scores on MMLU benchmark when using DeepSeek-R1-Distill-Qwen models both 14B and 8B
Here is the results
| Task | Version | Metric | Value | Stderr | |
|---|---|---|---|---|---|
| all | em | 0 | ± | 0 | |
| mmlu:abstract_algebra:0 | em | 0 | ± | 0 |
To Reproduce
Here is my bash script
MODEL=deepseek-ai/DeepSeek-R1-Distill-Qwen-14B
MODEL_ARGS="model_name=$MODEL,dtype=bfloat16,max_model_length=32768,gpu_memory_utilization=0.9,generation_parameters={max_new_tokens:32768,temperature:0.6,top_p:0.95}"
OUTPUT_DIR=lightevals_output_baseline/DeepSeek-R1-0528-Qwen3-14B/mml
TASK=mmlu
CUDA_VISIBLE_DEVICES=0,1 lighteval vllm $MODEL_ARGS "original|mmlu:abstract_algebra|0" \
--output-dir $OUTPUT_DIR \
--save-details
Here is one generated example:
{"doc":{"choices":[" A"," B"," C"," D"],"fewshot_samples":[],"fewshot_sorting_class":"False, False","generation_grammar":null,"generation_size":5,"gold_index":1,"id":"42","images":null,"instruction":"The following are multiple choice questions (with answers) about abstract algebra.\n\n","num_samples":1,"original_query":null,"query":"The following are multiple choice questions (with answers) about abstract algebra.\n\nQuestion: Statement 1 | Some abelian group of order 45 has a subgroup of order 10. Statement 2 | A subgroup H of a group G is a normal subgroup if and only if thenumber of left cosets of H is equal to the number of right cosets of H.\nA. True, True\nB. False, False\nC. True, False\nD. False, True\nAnswer:","sampling_methods":["GENERATIVE"],"specific":null,"stop_sequences":["\n"],"task_name":"mmlu:abstract_algebra|0","unconditioned_query":null,"use_logits":true},"metric":{"em":0},"model_response":{"argmax_logits_eq_gold":[],"input":"<｜begin▁of▁sentence｜><｜User｜>The following are multiple choice questions (with answers) about abstract algebra.\n\nQuestion: Statement 1 | Some abelian group of order 45 has a subgroup of order 10. Statement 2 | A subgroup H of a group G is a normal subgroup if and only if thenumber of left cosets of H is equal to the number of right cosets of H.\nA. True, True\nB. False, False\nC. True, False\nD. False, True\nAnswer:<｜Assistant｜>","input_tokens":[151643,151669,785,2701,525,5248,5754,4755,320,4197,11253,8,911,8115,46876,382,14582,25,21756,220,16,760,4329,668,69564,1874,315,1973,220,19,20,702,264,80115,315,1973,220,16,15,13,21756,220,17,760,362,80115,472,315,264,1874,479,374,264,4622,80115,421,323,1172,421,1221,890,315,2115,7960,1415,315,472,374,6144,311,279,1372,315,1290,7960,1415,315,472,624,32,13,3007,11,3007,198,33,13,3557,11,3557,198,34,13,3007,11,3557,198,35,13,3557,11,3007,198,16141,25,151670],"logits":null,"logprobs":[],"output_tokens":[[151667,198,40,1184,311,15442,1378,12239,911,8115,46876,323,8253,892,10601,315,830,476,895,374,4396,13,576,2606,525,362,13,3007,11,3007,26,425,13,3557,11,3557,26,356,13,3007,11,3557,26,422,13,3557,11,3007,382,5338,11,21756,220,16,25,330,8373,668,69564,1874,315,1973,220,19,20,702,264,80115,315,1973,220,16,15,2217,8636,220,17,25,330,32,80115,472,315,264,1874,479,374,264,4622,80115,421,323,1172,421,279,1372,315,2115,7960,1415,315,472,374,6144,311,279,1372,315,1290,7960,1415,315,472,2217,40,1184,311,1779,279,8046,315,1817,5114,382,24617,448,21756,220,16,382,4431,220,19,20,13,220,19,20,374,220,24,3039,220,20,11,773,220,18,61,17,353,220,20,13,8704,432,594,668,69564,11,358,1265,2908,279,3204,668,69564,5203,315,1973,220,19,20,382,1359,279,15811,57817,315,34226,668,69564,5203,11,279,668,69564,5203,315,1973,220,19,20,525,1447,12,1863,15159,19,20,92,21103,227,1863,62,24,24768,1863,62,20,271,12,1863,62,18,24768,1863,62,16,20,21103,227,1863,62,18,24768,1863,62,18,24768,1863,62,20,271,57,62,16,20,374,1863,62,18,24768,1863,62,20,11,773,1863,62,18,24768,1863,62,16,20,284,1863,62,18,24768,320,57,62,18,24768,1863,62,20,8,284,1863,62,18,24768,1863,62,18,24768,1863,62,20,382,3036,1863,15159,19,20,92,284,1863,62,24,24768,1863,62,20,382,7039,11,1863,62,24,24768,1863,62,20,702,1973,220,19,20,11,323,1863,62,18,24768,1863,62,18,24768,1863,62,20,1083,702,1973,220,19,20,382,7039,11,1558,2987,315,1493,614,264,80115,315,1973,220,16,15,1939,4431,220,16,15,374,220,17,9,20,382,3983,220,19,20,374,10322,11,773,902,2392,315,1973,220,17,11,2474,279,1874,374,668,69564,11,773,902,2392,702,1973,220,17,11,773,902,80115,315,1973,220,16,15,11,1576,220,16,15,7460,458,2392,315,1973,49702,220,16,15,11,714,220,17,3171,944,21749,279,27690,476,2494,382,641,458,668,69564,1874,11,421,1052,594,264,80115,315,1973,220,16,15,11,1221,553,14413,1387,88,594,57817,11,2474,220,17,323,220,20,21749,220,16,15,11,323,220,16,15,64828,220,19,20,30,2308,11,14413,1387,88,594,57817,2727,429,421,264,10250,64828,279,1973,315,279,1874,11,1221,1052,594,458,2392,315,429,10250,1973,382,3983,1588,11,279,1874,702,1973,220,19,20,11,892,374,74916,553,220,18,323,220,20,11,714,537,553,220,17,382,19,20,374,10322,11,773,902,2392,315,1973,220,17,382,54815,11,902,80115,315,1973,220,16,15,11,1576,220,16,15,374,1496,11,323,678,5424,614,10322,1973,2474,279,1874,1973,374,10322,382,641,458,668,69564,1874,11,279,1973,315,5424,64828,279,1874,1973,11,773,10163,525,3429,41214,315,220,19,20,11,892,525,220,16,11,18,11,20,11,24,11,16,20,11,19,20,11,678,10322,11,773,902,1496,1973,11,773,902,2392,315,1973,220,17,11,16085,902,80115,315,1973,220,16,15,382,3983,374,429,14016,1939,32,80115,315,1973,220,16,15,1035,1373,5424,315,1973,220,17,476,220,20,11,714,1973,220,20,374,3204,11,2474,220,20,64828,220,19,20,11,773,1052,525,5424,315,1973,220,20,382,3983,1973,220,16,15,7460,458,2392,315,1973,220,16,15,476,264,10601,382,3983,304,419,1142,11,2474,902,2392,315,1973,220,17,11,1496,421,1052,374,458,2392,315,1973,220,20,11,498,646,944,614,264,80115,7907,553,5424,315,1973,220,20,323,2494,11,714,2041,1973,220,17,11,498,646,944,614,1973,220,16,15,382,7661,23638,11,279,1973,315,279,80115,64828,279,1973,315,279,1874,11,773,1186,16753,646,614,1973,49702,220,19,20,11,773,3204,10163,525,220,16,11,18,11,20,11,24,11,16,20,11,19,20,382,16,15,1558,537,21749,220,19,20,11,773,553,32371,9669,594,57817,11,902,80115,315,1973,220,16,15,646,3000,304,264,1874,315,1973,220,19,20,11,1576,279,1973,1969,21749,279,1874,1973,382,43,351,9669,594,57817,2727,429,369,264,34226,1874,11,279,1973,315,894,80115,64828,279,1973,315,279,1874,382,9454,11,773,369,1973,220,19,20,11,1186,16753,1969,614,1973,49702,220,19,20,11,773,220,16,11,18,11,20,11,24,11,16,20,11,19,20,382,16,15,1558,537,21749,220,19,20,11,773,902,80115,315,1973,220,16,15,646,3000,304,894,1874,315,1973,220,19,20,11,668,69564,476,537,382,3983,279,5114,2727,330,14689,668,69564,1874,315,1973,220,19,20,497,714,2474,432,594,911,279,1874,1973,11,432,3171,944,4925,421,432,594,668,69564,476,537,26,902,80115,315,1973,220,16,15,304,894,1874,315,1973,220,19,20,382,3983,1077,752,7683,382,19,20,17779,553,220,16,15,374,220,19,13,20,11,537,7546,11,773,537,74916,382,9454,11,773,21756,220,16,374,895,382,3983,279,5114,2727,330,14689,668,69564,1874,497,714,2474,902,1874,315,1973,220,19,20,702,264,80115,315,1973,220,16,15,11,432,594,895,382,7039,11,374,1052,458,668,69564,1874,315,1973,220,19,20,30,7414,11,438,358,10007,11,714,807,1513,944,614,1186,16753,315,1973,220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need to evaluate two statements about abstract algebra and determine which combination of true or false is correct. The options are A. True, True; B. False, False; C. True, False; D. False, True.\n\nFirst, Statement 1: \"Some abelian group of order 45 has a subgroup of order 10.\"\n\nStatement 2: \"A subgroup H of a group G is a normal subgroup if and only if the number of left cosets of H is equal to the number of right cosets of H.\"\n\nI need to check the truth of each statement.\n\nStarting with Statement 1.\n\nOrder 45. 45 is 9 times 5, so 3^2 * 5. Since it's abelian, I should consider the possible abelian groups of order 45.\n\nBy the fundamental theorem of finite abelian groups, the abelian groups of order 45 are:\n\n- Z_{45} ≅ Z_9 × Z_5\n\n- Z_3 × Z_15 ≅ Z_3 × Z_3 × Z_5\n\nZ_15 is Z_3 × Z_5, so Z_3 × Z_15 = Z_3 × (Z_3 × Z_5) = Z_3 × Z_3 × Z_5.\n\nAnd Z_{45} = Z_9 × Z_5.\n\nNow, Z_9 × Z_5 has order 45, and Z_3 × Z_3 × Z_5 also has order 45.\n\nNow, does either of these have a subgroup of order 10?\n\nOrder 10 is 2*5.\n\nBut 45 is odd, so no element of order 2, since the group is abelian, so no element has order 2, so no subgroup of order 10, because 10 requires an element of order dividing 10, but 2 doesn't divide the exponent or something.\n\nIn an abelian group, if there's a subgroup of order 10, then by Cauchy's theorem, since 2 and 5 divide 10, and 10 divides 45? No, Cauchy's theorem says that if a prime divides the order of the group, then there's an element of that prime order.\n\nBut here, the group has order 45, which is divisible by 3 and 5, but not by 2.\n\n45 is odd, so no element of order 2.\n\nTherefore, no subgroup of order 10, because 10 is even, and all elements have odd order since the group order is odd.\n\nIn an abelian group, the order of elements divides the group order, so orders are divisors of 45, which are 1,3,5,9,15,45, all odd, so no even order, so no element of order 2, hence no subgroup of order 10.\n\nBut is that sufficient?\n\nA subgroup of order 10 would require elements of order 2 or 5, but order 5 is possible, since 5 divides 45, so there are elements of order 5.\n\nBut order 10 requires an element of order 10 or a combination.\n\nBut in this case, since no element of order 2, even if there is an element of order 5, you can't have a subgroup generated by elements of order 5 and something, but without order 2, you can't have order 10.\n\nMore precisely, the order of the subgroup divides the order of the group, so subgroups can have order dividing 45, so possible orders are 1,3,5,9,15,45.\n\n10 does not divide 45, so by Lagrange's theorem, no subgroup of order 10 can exist in a group of order 45, because the order must divide the group order.\n\nLagrange's theorem says that for a finite group, the order of any subgroup divides the order of the group.\n\nYes, so for order 45, subgroups must have order dividing 45, so 1,3,5,9,15,45.\n\n10 does not divide 45, so no subgroup of order 10 can exist in any group of order 45, abelian or not.\n\nBut the statement says \"some abelian group of order 45\", but since it's about the group order, it doesn't matter if it's abelian or not; no subgroup of order 10 in any group of order 45.\n\nBut let me confirm.\n\n45 divided by 10 is 4.5, not integer, so not divisible.\n\nYes, so Statement 1 is false.\n\nBut the statement says \"some abelian group\", but since no group of order 45 has a subgroup of order 10, it's false.\n\nNow, is there an abelian group of order 45? Yes, as I listed, but they don't have subgroups of order 10.\n\nPerhaps I should consider if there is a group where 10 divides 45, but no.\n\nUnless the group is not of order 45, but the statement is about groups of order 45.\n\nAnother thing: are there non-abelian groups of order 45? But the statement specifies abelian, but even if there were non-abelian, still no subgroup of order 10.\n\nBut for Statement 1, it's clear: no subgroup of order 10 in any group of order 45.\n\nSo Statement 1 is false.\n\nNow, Statement 2: \"A subgroup H of a group G is a normal subgroup if and only if the number of left cosets of H is equal to the number of right cosets of H.\"\n\nNow, the number of left cosets is [G:H], the index.\n\nSimilarly, number of right cosets is also [G:H], so they are always equal, right?\n\nIn any group, for a subgroup H, the number of left cosets equals the number of right cosets, both are [G:H].\n\nAnd [G:H] is defined as |G|/|H| if H is finite, but in general, it's the index.\n\nBut the point is, they are always equal, regardless of whether H is normal or not.\n\nIf H is normal, then left and right cosets coincide, but the number is still the same.\n\nThe statement says \"if and only if the number of left cosets equals the number of right cosets.\"\n\nBut since they are always equal, this condition is always true.\n\nTherefore, the \"if and only if\" is trivial because the premise is always true.\n\nSo, for any subgroup, the number of left and right cosets are equal.\n\nTherefore, the condition holds for all subgroups, but normal subgroups are a special case.\n\nBut the statement says \"if and only if\", meaning that H is normal iff the numbers are equal.\n\nBut since the numbers are always equal, the \"if\" part is true: if the numbers are equal (which they always are), then H is normal? No, not necessarily.\n\nLet's parse the logic.\n\nThe statement is: H is normal iff the number of left cosets equals number of right cosets.\n\nNow, the number of left cosets equals number of right cosets is always true, as established.\n\nTherefore, the \"if\" part: if the numbers are equal, then H is normal.\n\nBut this is not true, because there are non-normal subgroups where the numbers are still equal.\n\nFor example, take S3, the symmetric group on 3 letters.\n\nOrder 6.\n\nSubgroup H = {(1),(12)} which is order 2.\n\nNumber of left cosets: [S3:H] = 6/2 = 3.\n\nSimilarly, number of right cosets is 3.\n\nBut is H normal? (123)H(123)^{-1} = (123)(12)(132) = let's compute.\n\n(123)(12)(132): first (132) is inverse, (123)^{-1} = (132).\n\n(123)(12)(132): apply (132) first, then (12), then (123).\n\nBetter to think: conjugation: (123) * (12) * (132).\n\nSince (123)(132) = identity, no.\n\nConjugation: g h g^{-1}.\n\nSo (123) * (12) * (132).\n\n(123)(132) is not identity.\n\n(123)(132) = (1)(2)(3) no, permutation composition.\n\nAssume right action or left, but in S3, let's list elements.\n\nElements: (1), (12), (13), (23), (123), (132).\n\nH = {(1), (12)}\n\nNow, conjugate by (123): (123) * (12) * (132)\n\nFirst, (132) is (132), then (12), then (123).\n\nBut better to compute the conjugate: (123) * (12) * (132)\n\nSince (132) is inverse, but let's compute step by step.\n\n(123) * (12) = (13)(2) no, (123) sends 1 to 2, 2 to 3, 3 to 1.\n\n(12) sends 1 to 2, 2 to 1, 3 to 3.\n\nSo (123)(12): apply (12) first: if x, (12) x, then (123) of that.\n\nStandard is to apply right to left or left to right? In group theory, usually right action or left, but for permutations, often left to right or right to left.\n\nI think in S_n, composition is from left to right or right to left? Convention varies.\n\nLet's use cycle notation.\n\n(123) * (12): apply (12) first, then (123).\n\nSo for a point, say 1: (12) sends 1 to 2, then (123) sends 2 to 3, so 1 to 3.\n\n2: (12) sends 2 to 1, (123) sends 1 to 2, so 2 to 2? 2 to 1 to 2, so fixed.\n\nBetter: let σ = (123), τ = (12)\n\nThen σ τ (x) = σ(τ(x))\n\nSo for x=1: τ(1)=2, σ(2)=3, so 1 to 3\n\nx=2: τ(2)=1, σ(1)=2, so 2 to 2\n\nx=3: τ(3)=3, σ(3)=1, so 3 to 1\n\nSo σ τ = (13)\n\nSimilarly, (123) * (12) = (13)\n\nNow, conjugate: g h g^{-1} with g=(123), h=(12)\n\ng^{-1} = (132)\n\nSo g h g^{-1} = (123) * (12) * (132)\n\nNow, (123) * (12) = (13) as above.\n\nThen (13) * (132)\n\n(13) * (132): apply (132) first, then (13)\n\nx=1: (132) sends 1 to 3, (13) sends 3 to 1, so 1 to 1\n\nx=2: (132) sends 2 to 1, (13) sends 1 to 3, so 2 to 3\n\nx=3: (132) sends 3 to 2, (13) sends 2 to 2, so 3 to 2\n\nSo (2 3)\n\nThus, (123) * (12) * (132) = (23)\n\nWhich is not in H, since H has (1) and (12), (23) is different.\n\nSo H is not normal.\n\nNow, number of left cosets: [S3:H] = 3, similarly right cosets 3, equal.\n\nBut H not normal.\n\nTherefore, the number being equal does not imply normal.\n\nNow, the other direction: if H is normal, then certainly the numbers are equal, but since they are always equal, it's true, but the iff is broken because the converse is false.\n\nIn the statement, it says \"if and only if\", so both directions must hold.\n\nFirst direction: if normal, then numbers equal. True, because if normal, cosets are the same, so same number.\n\nSecond direction: if numbers equal, then normal. But as above, false.\n\nTherefore, Statement 2 is false.\n\nBecause the \"only if\" part is not true; non-normal subgroups can have the same number of left and right cosets.\n\nIn general, for any subgroup, the number of left and right cosets are equal, so the condition is always true, but normal is not always true.\n\nHence, the statement is false.\n\nNow, back to Statement 1, which we said is false.\n\nSo both statements are false.\n\nTherefore, answer should be B. False, False.\n\nBut let me double-check Statement 1.\n\nStatement 1: \"Some abelian group of order 45 has a subgroup of order 10.\"\n\nAs I thought, no, because 10 does not divide 45.\n\nBut is there an abelian group of order 45 that has a subgroup of order 10? No, because order must divide.\n\nUnless the group is infinite, but it's finite, order 45.\n\nPerhaps I misread the order.\n\nOrder 45, subgroup order 10.\n\n10 does not divide 45.\n\nMaybe it's order 30 or something, but no.\n\nAnother thing: in some contexts, but no, Lagrange's theorem applies.\n\nPerhaps for abelian groups, but no, Lagrange is for all finite groups.\n\nSo Statement 1 false.\n\nStatement 2 false.\n\nSo B. False, False.\n\nBut let me see the options.\n\nA. True, True\n\nB. False, False\n\nC. True, False\n\nD. False, True\n\nSo B.\n\nBut in Statement 2, is there any case where the numbers are not equal?\n\nIn general, for groups, the number of left and right cosets are always equal, so the \"if and only if\" is incorrect because the premise is always true.\n\nIf the group were such that left and right cosets could differ, but in standard group theory, they don't.\n\nFor example, in non-abelian groups, left and right cosets can be different, but the number is still the same.\n\nYes.\n\nUnless the subgroup is not defined, but in general, for finite groups, [G:H] is defined and equal.\n\nSo yes.\n\nTherefore, both false.\n\nSo answer should be B.\n\nBut the question is to box the answer.\n\nBut let me see the original question.\n\n\"Answer:\" and then I need to provide the choice.\n\nPerhaps I should write it.\n\nBut in the initial, it says \"Answer:\" and then the options.\n\nBut I think B is correct.\n\nBut let me see if Statement 1 could be true.\n\nPerhaps there is a mistake.\n\nAnother thought: maybe \"some abelian group of order 45\" but 45 has prime factors, but no.\n\nPerhaps in the context, but I think not.\n\nMaybe it's about cyclic group or something, but no.\n\nSo I think Statement 1 is false.\n\nSimilarly, Statement 2 is false.\n\nSo B. False, False.\n</think>\nStatement 1 is false because no group of order 45 can have a subgroup of order 10, as the order of any subgroup must divide the group order by Lagrange's theorem, and 10 does not divide 45. This holds for both abelian and non-abelian groups.\n\nStatement 2 is false because the number of left and right cosets of any subgroup H in a group G is always equal, regardless of whether H is normal. The condition that the numbers are equal is always true, but it does not imply that H is normal, as shown by the example of the subgroup H = {(1), (12)} in S3, which has three left and three right cosets but is not normal.\n\nTherefore, both statements are false, and the correct option is B.\n\n\\boxed{\\text{B}}"],"text_post_processed":["\nStatement 1 is false because no group of order 45 can have a subgroup of order 10, as the order of any subgroup must divide the group order by Lagrange's theorem, and 10 does not divide 45. This holds for both abelian and non-abelian groups.\n\nStatement 2 is false because the number of left and right cosets of any subgroup H in a group G is always equal, regardless of whether H is normal. The condition that the numbers are equal is always true, but it does not imply that H is normal, as shown by the example of the subgroup H = {(1), (12)} in S3, which has three left and three right cosets but is not normal.\n\nTherefore, both statements are false, and the correct option is B.\n\n\\boxed{\\text{B}}"],"truncated_tokens_count":0,"unconditioned_logprobs":null}}
Seems like the generated answer matches the gold answer but the em score is 0. Is there something wrong with the parser?
Version info
Name: lighteval
Version: 0.12.1