A Simple Gap-producing Reduction for the Parameterized Set Cover Problem

Abstract: Given an $n$-vertex bipartite graph $I=(S,U,E)$, the goal of set cover problem is to find a minimum sized subset of $S$ such that every vertex in $U$ is adjacent to some vertex of this subset. It is NP-hard to approximate set cover to within a $(1-o(1))\ln n$ factor. If we use the size of the optimum solution $k$ as the parameter, then it can be solved in $n<sup>{k+o(1)}$ time. A natural question is: can we approximate set cover to within an $o(\ln n)$ factor in $n<sup>{k-\epsilon}$ time? In a recent breakthrough result, Karthik, Laekhanukit and Manurangsi showed that assuming the Strong Exponential Time Hypothesis (SETH), for any computable function $f$, no $f(k)\cdot n<sup>{k-\epsilon}$-time algorithm can approximate set cover to a factor below $(\log n)<sup>{\frac{1}{poly(k,e(\epsilon))}}$ for some function $e$. This paper presents a simple gap-producing reduction which, given a set cover instance $I=(S,U,E)$ and two integers $k&lt;h\le (1-o(1))\sqrt[k]{\log |S|/\log\log |S|}$, outputs a new set cover instance $I&#39;=(S,U&#39;,E&#39;)$ with $|U&#39;|=|U|<sup>{h<sup>k}|S|<sup>{O(1)}$ in $|U|^{{h^k}\cdot} |S|<sup>{O(1)}$ time such that: (1) if $I$ has a $k$-sized solution, then so does $I&#39;$; (2) if $I$ has no $k$-sized solution, then every solution of $I&#39;$ must contain at least $h$ vertices. Setting $h=(1-o(1))\sqrt[k]{\log |S|/\log\log |S|}$, we show that assuming SETH, for any computable function $f$, no $f(k)\cdot n<sup>{k-\epsilon}$-time algorithm can distinguish between a set cover instance with $k$-sized solution and one whose minimum solution size is at least $(1-o(1))\cdot \sqrt[k]{\frac{\log n}{\log\log n}}$. This improves the result of Karthik, Laekhanukit and Manurangsi.