In this paper, we prove that assuming the exponential time hypothesis (ETH), there is no $f(k)\cdot n^{k^{o(1/\log\log k)}}$-time algorithm that can decide whether an $n$-vertex graph contains a clique of size $k$ or contains no clique of size $k/2$, and no FPT algorithm can decide whether an input graph has a clique of size $k$ or no clique of size $k/f(k)$, where $f(k)$ is some function in $k^{1-o(1)}$. Our results significantly improve the previous works [Lin21, LRSW22]. The crux of our proof is a framework to construct gap-producing reductions for the \kclique{} problem. More precisely, we show that given an error-correcting code $C:\Sigma_1^k\to\Sigma_2^{k'}$ that is locally testable and smooth locally decodable in the parallel setting, one can construct a reduction which on input a graph $G$ outputs a graph $G'$ in $(k')^{O(1)}\cdot n^{O(\log|\Sigma_2|/\log|\Sigma_1|)}$ time such that: $\bullet$ If $G$ has a clique of size $k$, then $G'$ has a clique of size $K$, where $K = (k')^{O(1)}$. $\bullet$ If $G$ has no clique of size $k$, then $G'$ has no clique of size $(1-\varepsilon)\cdot K$ for some constant $\varepsilon\in(0,1)$. We then construct such a code with $k'=k^{\Theta(\log\log k)}$ and $|\Sigma_2|=|\Sigma_1|^{k^{0.54}}$, establishing the hardness results above. Our code generalizes the derivative code [WY07] into the case with a super constant order of derivatives.
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在本文中,我们证明在假设指数时间假设(ETH)的情况下,不存在$f(k)\cdot n^{k^{o(1/\log\log k)}}$时间的算法,该算法可以确定$n$顶点图是否包含大小为$k$或是否不包含大小为$k/2$的大派系,并且没有FPT算法可以确定输入图是否具有大小为$k$的大派系或是否不具有大小为$k/f(k)$的大派系,其中$f(k)$是$k^{1-o(1)}$中的某个函数。我们的结果显着改进了以前的工作[Lin21,LRSW22]。我们证明的关键是构造具有间隙产生效果的约简框架以解决\kclique{}问题。更确切地说,我们展示了如何利用在并行环境中是本地可测试和平滑本地可解码的纠错码$C:\Sigma_1^k\to\Sigma_2^{k'}$来构造一个规约夫$G$输入图输出图$G'$,此规约时间复杂度是$(k')^{O(1)}\cdot n^{O(\log|\Sigma_2|/\log|\Sigma_1|)}$,满足如下条件:$\bullet$如果$G$有一个大小为$k$的派系,那么$G'$有一个大小为$K$的派系,其中$K=(k')^{O(1)}$。$\bullet$如果$G$没有大小为$k$的派系,则$G'$没有大小为$(1-\varepsilon)\cdot K$的派系,其中$\varepsilon\in(0,1)$是一些常数。我们构造的这种代码具有$k'=k^{\Theta(\log\log k)}$和$|\Sigma_2|=|\Sigma_1|^{k^{0.54}}$,从而确定了以上难度结果。我们的代码将导数码[WY07]推广到具有超常导数阶的情况。