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 [I. Dinur and D. Steurer, 2014]. If we use the size of the optimum solution k as the parameter, then it can be solved in n^{k+o(1)} time [Eisenbrand and Grandoni, 2004]. 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 et al., 2018], Karthik, Laekhanukit and Manurangsi showed that assuming the Strong Exponential Time Hypothesis (SETH), for any computable function f, no f(k)* n^{k-epsilon}-time algorithm can approximate set cover to a factor below (log n)^{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 <=(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}* |S|^{O(1)} time such that 
- if I has a k-sized solution, then so does I'; 
- 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)* 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))* sqrt[k]((log n)/(log log n)). This improves the result in [Karthik et al., 2018].