如果在强指数时间假设下边着色问题很难，那么强指数时间假设就是错误的
If Edge Coloring is Hard under SETH, then SETH is False
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中文总结 AI 辅助
探讨能否从SAT、3 - SUM或APSP的下界推出边着色问题的\(2^{\Omega(n^2)}\)下界来解释其上界缺失，给出否定答案，即若在相关假设下有这样的归约，相应假设就是错误的。
中文摘要 AI 辅助
边着色问题非常困难：对于输入图有\(n\)个节点的情况，能否在\(2^{o(n^2)}\)（更不用说\(2^{O(n)}\)）时间内解决仍是未知。能否通过从SAT、3 - SUM或APSP的下界推导出\(2^{\Omega(n^2)}\)的下界来解释这种上界的缺失？本文给出否定答案：如果有一种归约表明在上述问题之一的已知算法最优的假设下，边着色不能比\(\alpha^{n^2}\)（\(\alpha>1\)是明确常数）更快解决，那么相应假设就是错误的。
英文摘要
The Edge Coloring problem is notoriously hard: it is still unknown whether it can be solved in time $2^{o(n^2)}$ (let alone $2^{O(n)}$), where $n$ is the number of nodes of the input graph. Can one explain the lack of such upper bounds by deriving a lower bound $2^{Ω(n^2)}$ from a lower bound for SAT, $3$-SUM, or APSP? In this note, we provide a negative answer for this question: if there is a reduction showing that Edge Coloring cannot be solved faster than in $α^{n^2}$ (where $α>1$ is an explicit constant) under a hypothesis that known algorithms for one of the problems mentioned above are optimal, then the corresponding hypothesis is false.
