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

## TL;DR

This paper introduces a simple reduction technique for the parameterized set cover problem that enhances existing hardness results, showing that under SETH, approximating within certain factors in sub-exponential time is infeasible.

## Contribution

It presents a new gap-producing reduction for set cover that improves lower bounds on approximation hardness under SETH.

## Key findings

- Establishes a reduction that amplifies the gap between yes and no instances.
- Shows that no algorithm can distinguish certain set cover instances within specified time bounds.
- Improves upon previous hardness results by Karthik, Laekhanukit, and Manurangsi.

## 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 $n^{k+o(1)}$ time. A natural question is: can we approximate set cover to within an $o(\ln n)$ factor in $n^{k-\epsilon}$ 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 $f(k)\cdot n^{k-\epsilon}$-time algorithm can approximate set cover to a factor below $(\log n)^{\frac{1}{poly(k,e(\epsilon))}}$ 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 $k<h\le (1-o(1))\sqrt[k]{\log |S|/\log\log |S|}$, outputs a new set cover instance $I'=(S,U',E')$ with $|U'|=|U|^{h^k}|S|^{O(1)}$ in $|U|^{h^k}\cdot |S|^{O(1)}$ time such that: (1) if $I$ has a $k$-sized solution, then so does $I'$; (2) if $I$ has no $k$-sized solution, then every solution of $I'$ 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 $f(k)\cdot n^{k-\epsilon}$-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.