{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"scicode","formal_name":"SciCode","introduction":"SciCode evaluates the ability to solve scientific research problems through code. Problems are decomposed into subproblems; this dev import preserves the relationships between 15 parent problems and 50 subproblems.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/SciCode1/SciCode","indexing_mode":"noindex","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[]}},"task_id":"f012d95b-9559-50e1-98f2-c1c7f2657779","task_key":"dev--96fdd33e-7a8f-5191-a8cf-e8ce0b324581--38~2e1","task_revision_id":"3","upstream_id":"38.1","short_description":"Given two vectors, return the cross-product of these two vectors. The input…","config":"","split":"dev","body":"{\"step_background\":\"Background\\nGiven the two input vectors\\n$$\\n\\\\begin{aligned}\\n& \\\\mathbf{a}=a_1 \\\\mathbf{i}+a_2 \\\\mathbf{j}+a_3 \\\\mathbf{k} \\\\\\\\\\n& \\\\mathbf{b}=b_1 \\\\mathbf{i}+b_2 \\\\mathbf{j}+b_3 \\\\mathbf{k}\\n\\\\end{aligned}\\n$$\\ntheir cross product a × b can be expanded using distributivity:\\n$$\\n\\\\begin{aligned}\\n\\\\mathbf{a} \\\\times \\\\mathbf{b}= & \\\\left(a_1 \\\\mathbf{i}+a_2 \\\\mathbf{j}+a_3 \\\\mathbf{k}\\\\right) \\\\times\\\\left(b_1 \\\\mathbf{i}+b_2 \\\\mathbf{j}+b_3 \\\\mathbf{k}\\\\right) \\\\\\\\\\n= & a_1 b_1(\\\\mathbf{i} \\\\times \\\\mathbf{i})+a_1 b_2(\\\\mathbf{i} \\\\times \\\\mathbf{j})+a_1 b_3(\\\\mathbf{i} \\\\times \\\\mathbf{k})+ \\\\\\\\\\n& a_2 b_1(\\\\mathbf{j} \\\\times \\\\mathbf{i})+a_2 b_2(\\\\mathbf{j} \\\\times \\\\mathbf{j})+a_2 b_3(\\\\mathbf{j} \\\\times \\\\mathbf{k})+ \\\\\\\\\\n& a_3 b_1(\\\\mathbf{k} \\\\times \\\\mathbf{i})+a_3 b_2(\\\\mathbf{k} \\\\times \\\\mathbf{j})+a_3 b_3(\\\\mathbf{k} \\\\times \\\\mathbf{k})\\n\\\\end{aligned}\\n$$\\n\\nFrom this decomposition, by using the above-mentioned equalities and collecting similar terms, we obtain:\\n\\n$$\\n\\\\begin{aligned}\\n\\\\mathbf{a} \\\\times \\\\mathbf{b}= & a_1 b_1 \\\\mathbf{0}+a_1 b_2 \\\\mathbf{k}-a_1 b_3 \\\\mathbf{j} \\\\\\\\\\n& -a_2 b_1 \\\\mathbf{k}+a_2 b_2 \\\\mathbf{0}+a_2 b_3 \\\\mathbf{i} \\\\\\\\\\n& +a_3 b_1 \\\\mathbf{j}-a_3 b_2 \\\\mathbf{i}+a_3 b_3 \\\\mathbf{0} \\\\\\\\\\n= & \\\\left(a_2 b_3-a_3 b_2\\\\right) \\\\mathbf{i}+\\\\left(a_3 b_1-a_1 b_3\\\\right) \\\\mathbf{j}+\\\\left(a_1 b_2-a_2 b_1\\\\right) \\\\mathbf{k}\\n\\\\end{aligned}\\n$$\\n\\nFor column vectors, we can represent the same result as follows:\\n\\n$$\\n\\\\mathbf{a} \\\\times \\\\mathbf{b}=\\\\left[\\\\begin{array}{l}\\na_2 b_3-a_3 b_2 \\\\\\\\\\na_3 b_1-a_1 b_3 \\\\\\\\\\na_1 b_2-a_2 b_1\\n\\\\end{array}\\\\right]\\n$$\",\"step_description_prompt\":\"Given two vectors, return the cross-product of these two vectors. The input should be two numpy arrays and the output should be one numpy array.\"}","display_format":"scicode-step","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/SciCode1/SciCode","history":"initial import","indexing_mode":"noindex","subproblems":[],"grids":[]}