初始化项目，由ModelHub XC社区提供模型

Model: nv-community/Nemotron-Cascade-8B
Source: Original Platform

evaluation/data/mmlu/flan_cot_fewshot/mmlu_abstract_algebra.yaml

@@ -0,0 +1,59 @@
+dataset_name: abstract_algebra
+description: The following are multiple choice questions (with answers) about abstract
+  algebra.
+fewshot_config:
+  sampler: first_n
+  samples:
+  - question: 'Statement 1 | Every element of a group generates a cyclic subgroup of
+      the group. Statement 2 | The symmetric group S_10 has 10 elements.
+
+      (A) True, True (B) False, False (C) True, False (D) False, True'
+    target: Let's think step by step. A cyclic group is a group that is generated
+      by a single element. Hence a subgroup generated by a single element of a group
+      is cyclic and Statement 1 is True. The answer is (C).
+  - question: 'The symmetric group $S_n$ has $
+
+      actorial{n}$ elements, hence it is not true that $S_{10}$ has 10 elements.
+
+      Find the characteristic of the ring 2Z.
+
+      (A) 0 (B) 3 (C) 12 (D) 30'
+    target: Let's think step by step. A characteristic of a ring is R is $n$ if the
+      statement $ka = 0$ for all $a\in 2Z$ implies that $k$ is a multiple of $n$.
+      Assume that $ka = 0$ for all $a\in 2Z$ for some $k$. In particular $2k = 0$.
+      Hence $k=0$ and $n=0$. The answer is (A).
+  - question: 'Statement 1| Every function from a finite set onto itself must be one
+      to one. Statement 2 | Every subgroup of an abelian group is abelian.
+
+      (A) True, True (B) False, False (C) True, False (D) False, True'
+    target: "Let's think step by step. Statement 1 is true. Let $S$ be a finite set.\
+      \ If $f:S \nightarrow S$ is a onto function, then $|S| = |f(S)|$. If $f$ was\
+      \ not one to one, then for finite domain $S$ the image would have less than\
+      \ $S$ elements, a contradiction.\nStatement 2 is true. Let $G$ be an abelian\
+      \ group and $H$ be a subgroup of $G$. We need to show that $H$ is abelian. Let\
+      \ $a,b \\in H$. Then $a,b \\in G$ and $ab=ba$. Since $G$ is abelian, $ab=ba$.\
+      \ Since $H$ is a subgroup of $G$, $ab \\in H$. Therefore, $ab=ba$ and $H$ is\
+      \ abelian. The answer is (A)."
+  - question: 'Statement 1 | If aH is an element of a factor group, then |aH| divides
+      |a|. Statement 2 | If H and K are subgroups of G then HK is a subgroup of G.
+
+      (A) True, True (B) False, False (C) True, False (D) False, True'
+    target: Let's think step by step. Statement 2 is false. Let $H$ be a subgroup
+      of $S_3$ generated by the cycle $(1,2)$ and $K$ be a subgroup of $S_3$ generated
+      by the cycle $(1,3)$. Both $H$ and $K$ have two elements, the generators and
+      the identity. However $HK$ contains cycles (1,2), (1,3) and (2,3,1), but the
+      inverse of (2,3,1) is (2,1,3) and it does not belong to HK, hence HK is not
+      a subgroup. The answer is (B).
+  - question: 'Find all c in Z_3 such that Z_3[x]/(x^2 + c) is a field.
+
+      (A) 0 (B) 1 (C) 2 (D) 3'
+    target: 'Let''s think step by step. Z_3[x]/(x^2 + c) is a field if and only if
+      x^2 + c does not have roots in Z_3. That is x^2 + c != 0 for every x in Z_3.
+      If c = 0, then x^2 + c = x^2 has root 0. If c = 1 then x^2 + c = x^2 + 1 = 0
+      + 1 for x = 0, 1 + 1 = 2 for x = 1 and 1 + 1 = 2 for x = 2, hence x^2 + 1 does
+      not have any roots. For c = 2 the polynomial x^2 + 2 has two roots at x = 1
+      and x = 2. Hence Z_3[x]/(x^2 + c) is a field if and only if c = 1. The answer
+      is (B).'
+tag: mmlu_flan_cot_fewshot_stem
+include: _mmlu_flan_cot_fewshot_template_yaml
+task: mmlu_flan_cot_fewshot_abstract_algebra