展示 HN：如何获得雅可比猜想反驳的寓言 CoT
4 分•作者: SonOfLilit•3 个月前
自雅可比猜想反例发布以来，许多人要求发布一个思维链（Chain of Thought），以便他们能更好地理解导致反例的数学直觉。<p>由于没有发布，我尝试“密室逆向工程”它：给一个 Claude Fable（一个AI模型）结果，然后让它生成一个关于如何得出该结果的说明，但不能有任何剧透；接着让第二个 Fable 遵循这个说明，如果太容易就删除细节，如果太难就添加提示。我本可以进一步简化，但我停下来是因为该睡觉了，并且我分享出来是因为它让我对正在发生的事情有了一些直觉。<p>以下是一个成功运行的例子（可惜 Anthropic 隐藏了思维链）：<p><a href="https://claude.ai/share/80526d56-1c23-407d-8f5c-59a704221454" rel="nofollow">https://claude.ai/share/80526d56-1c23-407d-8f5c-59a704221454</a><p>## 直觉<p>这是我对其中有趣想法的理解（可能过于简略，Fable 难以在没有良好引导的情况下持续获得；姑且听之，我不是数学家；稍后有一个经过测试的完整提示，其中包含所有必要信息）：<p>* 避免使用 Bass–Connell–Wright 或 Drużkowski 标准形式（这些是重新参数化，在此类搜索中感觉很自然，但它们会通过权衡次数与维度来将低复杂度示例变成高复杂度示例，而反例是低复杂度的）
* 在 C^3 中寻找，而不是 C^2，C^2 可能不够有趣
* 寻找一个 3:1 的覆盖，而不是 2:1，欧拉（一如既往）有一个结果表明 2:1 不会奏效
* 寻找两个函数的组合（多项式比和剪切变换），它们在雅可比行列式处处为 `x` 和 `c/x`，除了在 `x=0` 处（使用标准技巧来处理未定义在某个点上的情况，并将所有问题推入那个点，这就是 `1 + xy` 的由来）<p># 重现思维链的提示<p>在此处： <a href="https://gist.github.com/SonOfLilit/8882a145048ba260b160568ba6f48093#prompt-for-reproducing-cot" rel="nofollow">https://gist.github.com/SonOfLilit/8882a145048ba260b160568ba...</a>
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Since the publication of the Jacobian Conjecture refutation, many people have asked for a Chain of Thought to be published so they can better understand the mathematical intuition that leads to the counterexample.<p>Since none was published, I tried to "clean room reverse engineer" it: give one Claude Fable the result and ask it to generate a writeup of how to arrive at it without any spoilers, then ask a second Fable to follow the write up, if it's too easy remove details, if it's too hard add hints. I could have simplified further, I stopped because it's time to sleep, and I'm sharing because it gave me some intuition for what's going on.<p>This is what a successful run looks like (sadly Anthropic hide Chains of Thought):<p><a href="https://claude.ai/share/80526d56-1c23-407d-8f5c-59a704221454" rel="nofollow">https://claude.ai/share/80526d56-1c23-407d-8f5c-59a704221454</a><p>## Intuition<p>This is my understanding of the interesting ideas (probably too summarized for Fable to consistently get it without a good harness; take with a grain of salt, I'm not a mathematician; later there is a full prompt that was tested and contains everything needed):<p>* No Bass–Connell–Wright or Drużkowski normal forms (those are reparametrizations that feel natural in this search but they turn low-complexity examples into high-complexity by trading off degree vs dimension, and the counterexample is low complexity)
* Look in C^3, not C^2, C^2 is probably not interesting enough
* Look for a 3:1 cover, not 2:1, there's some result due to Euler (as always) that shows 2:1 will not work
* Look for a composition of two functions (a ratio of polynomials and a shear) that have Jacobian determinants `x` and `c/x` everywhere, except at `x=0` (do the standard trick for not defining at a hole and shoving all the problems into that one hole, that's where the `1 + xy` comes from)<p># Prompt for reproducing CoT<p>Here: <a href="https://gist.github.com/SonOfLilit/8882a145048ba260b160568ba6f48093#prompt-for-reproducing-cot" rel="nofollow">https://gist.github.com/SonOfLilit/8882a145048ba260b160568ba...</a>