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XOR：读不出来，不一定是没留下 · 实现代码","sourceSlides":[],"knowledgeBlock":null,"formulas":[],"previousBlock":"b-d036dca29a622a26eaa66752","nextBlock":"b-2a4f663cf8ea2970c447f86c","prerequisites":[{"id":"b-adf8e2cd20e024db00882983","url":"#b-adf8e2cd20e024db00882983","role":"section-title"},{"id":"b-50385c867ba82bd715d190a2","url":"#b-50385c867ba82bd715d190a2","role":"problem-context"}],"relatedBlocks":[]},{"id":"b-2a4f663cf8ea2970c447f86c","type":"pre","scope":"document","url":"https://codingai-lec03.pages.dev/course/notebook-blocks/18ab1e7467c0443890a83ed274ee0499.html#b-2a4f663cf8ea2970c447f86c","content":"```\n# 独立 CPU XOR 检查；这里的变换是手工构造。\nh = [(-1,-1),(-1,1),(1,-1),(1,1)]\ny = [0,1,1,0]\nz = [a*b for a,b in h]\nprediction = [int(v<0) for v in z]\nassert prediction == y\nassert tuple((h[0][j]+h[3][j])/2 for j in (0,1)) == (0,0)\nassert tuple((h[1][j]+h[2][j])/2 for j in (0,1)) == 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