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.

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.

Equations

$$k(1+\ln |U^{\prime }|)=k(1+\ell \ln |U|+\ln m)<h,$$

$$A_i=\{(a_1,a_2,\dots ,a_k):\text{for all j∈[k], aj∈[h]}\}.$$

$${\vec{a}}_i=(M_{i,b_1},M_{i,b_2},\dots ,M_{i,b_k}).$$

$$M_{r,b}=M_{r,b^{\prime }}={\vec{a}}_r[j].$$

$$m\log \ell +mk\log p>-c\log n$$

$$m\log \ell +h\log n+m(k+1)\log p+m(k+1)\log h<0.$$

$$\frac{c\log n}{mk}+\frac{\log \ell }{k}>\frac{\log \ell }{(k+1)}+\frac{h\log n}{m(k+1)}+\log h.$$

$$|S(I)|+|U(I)|$$

$$\le M+M\log \log M\cdot (2Ck\log M{)}^{\frac{\log M}{(1+\delta /2{)}^k\log \log M}}$$

$$\le M+M\log \log M\cdot (\log M{)}^{\frac{2\log \log M}{(1+\delta /2{)}^k\log \log M}}\text{(using logM≥2Ck for large n)}$$

$$\le M+M\log \log M\cdot M^{\frac{2}{(1+\delta /2{)}^k}}$$

$$\le M+M\log \log M\cdot M^{1/k^4}\text{(using (1+δ/2)k≥2k4)}$$

$$\le M^{1+1/k^3}\text{(using M1/k3≥1+M1/k4loglogM for large n)}$$

$$=N.$$

$$\sum _{j=1}^k{x}_j=t\iff \exists i\in [s]\text{ such that }\sum _{j=1}^k{f}_i(x_j,t)=\vec{0}.$$

$$({\vec{x}}_1[j],{\vec{x}}_2[j],\dots ,{\vec{x}}_k[j])={\vec{a}}_{\ell }\in D\text{ for some ℓ∈[∣D∣]}.$$

$$\sum _{i\in [k]}{\vec{x}}_i[j]\ne 0.$$

$$({\vec{x}}_1[j],{\vec{x}}_2[j],\dots ,{\vec{x}}_k[j])\notin D.$$

$${\vec{x}}_{i_{\ell }}[j]\ne {\vec{a}}_{\ell }[i_{\ell }].$$

$$\vec{u}[\ell ]=i_{\ell }.$$

$$|S(I_i)|+|U(I_i)|\le kn+n^{1+1/k^3}\le N,$$

$$f(k)\cdot N^{\lceil k/2\rceil -ϵ}\le n^{\lceil k/2\rceil -ϵ+1/k},$$

$$\frac{1}{(1+\delta )}{\left(\frac{\log N}{\log \log N}\right)}^{1/k}\le \frac{1}{(1+\delta /2)}{\left(\frac{\log n}{\log \log n}\right)}^{1/k}.$$

$$v_i\in V_i$$

$$X_{\{i,j\}}=\{\{v_i,v_j\}:v_i\in V_i,v_j\in V_j,\{v_i,v_j\}\in X\}.$$

$$\{v_i\}=\bigcap _{j\in [k]\setminus \{i\},e\in X_{\{i,j\}}}e.$$

$${\frac{\log m}{\log \log m}}^{1/\left(\genfrac{}{}{0ex}{}{k}{2}\right)}\ge {\frac{\log N}{O(\log \log N)}}^{1/\left(\genfrac{}{}{0ex}{}{k}{2}\right)}\ge (\log N{)}^{1/ϵ(k^{\prime })}.$$

Full text

In a recent breakthrough result [24], 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 $k<h\leq(1-o(1))\sqrt[k]{\log|S|/\log\log|S|}$, outputs a new set cover instance $I^{\prime}=(S,U^{\prime},E^{\prime})$ with $|U^{\prime}|=|U|^{h^{k}}|S|^{O(1)}$ in $|U|^{h^{k}}\cdot|S|^{O(1)}$ time such that

• •

if $I$ has a $k$-sized solution, then so does $I^{\prime}$;

• •

if $I$ has no $k$-sized solution, then every solution of $I^{\prime}$ 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 in [24].

1 Introduction

We consider the set cover problem (SetCover): given an $n$-vertex bipartite graph $I=(S,U,E)$, where $U$ is the underlying universe set and $S$ represents the set family, find a minimum sized subset $C$ of $S$ such that every vertex of $U$ is adjacent to some vertex of $C$. We use $S(I)$, $U(I)$ and $opt(I)$ to denote the sets $S$, $U$ and the minimum size of the solution of $I$ respectively. A vertex $u\in U$ is covered by a subset $C\subseteq S$ if $u$ is adjacent to some vertex of $C$. The set cover problem is NP-hard [23]. Unless $P=NP$, we do not expect to solve it in polynomial time. One way to handle NP-hard problems is to use approximation algorithms. An algorithm of SetCover achieves an $r$-approximation if for every input instance $I$, it returns a subset $C$ of $S(I)$ such that $C$ covers $U(I)$ and $|C|\leq r\cdot opt(I)$. The polynomial time approximability of SetCover is well-understood: the greedy algorithm can output a solution of size at most $opt(I)\cdot(1+\ln n)$ [10, 21, 28, 34, 35] and it was shown that no polynomial time algorithm can achieve an approximation factor within $(1-o(1))\ln n$ unless $P=NP$ [4, 14, 17, 29, 32]. On the other hand, if we take the optimum solution size $k=opt(I)$ as a parameter, then the simple brute-force searching algorithm can solve this problem in $n^{k+1}$ time. Assuming the exponential time hypothesis (ETH) [19, 20], i.e., $3$-SAT on $n$ variables cannot be solved in $2^{o(n)}$ time, there is no $n^{o(k)}$ time algorithm for SetCover. Under the strong exponential time hypothesis (SETH) [19, 20], which claims that for any $\epsilon\in(0,1)$ there exists a $d\geq 3$ such that $d$-SAT on $n$ variables cannot be solved in $2^{(1-\epsilon)n}$ time, we can further rule out $n^{k-\epsilon}$-time algorithm for set cover for any $\epsilon>0$ [31]. It is quite natural to ask [11]:

Is there any $o(\ln n)$-approximation algorithm for the parameterized set cover problem (or dominating set problem) with running time $n^{k-\epsilon}$?

Exponential time approximation algorithms for the unparameterised version of set cover problem were studied in [7, 13]. It was shown that for any ratio $r$, there is a $(1+\ln r)$-approximation algorithm for SetCover with running time $2^{n/r}n^{O(1)}$. No $n^{k-\epsilon}$ time algorithm for SetCover achieving an approximation ratio in $o(\ln n)$ is known in literature. On the other hand, proving inapproximability for a parameterized problem is not an easy task. In fact, even the constant FPT-approximability, i.e., the existence of $f(k)\cdot n^{O(1)}$-time algorithm for any computable function $f$ (henceforth referred to as FPT-algorithm) with constant approximation, has been open for many years [30]. Lacking techniques like PCP-theorem [5], many results on the parameterized inapproximability of set cover problem had to use strong conjectures [6, 8] to create a gap in the first place. It is of great interest to develop techniques to prove hardness of approximation for parameterized problems only using hypothesis such as $SETH$, $ETH$ or even weaker assumptions like \text{\textup{W}[1]}\neq FPT or \text{\textup{W}[2]}\neq FPT [15, 18] from the parameterized complexity theory. The success of this quest might extend the arsenal of methods for proving hardness of approximation and lead to PCP-like theorems for Fine-Grained Complexity [3].

The first constant FPT-inapproximability result for parameterized SetCover based on \text{\textup{W}[1]}\neq FPT was given by [9] using the one-sided gap of Biclique from [26]. In fact, [9] deals with dominating set problem, which is essentially the same as SetCover. Recently, Karthik, Laekhanukit and Manurangsi [24] significantly improved the FPT-inapproximation factor to $(\log n)^{1/{k^{O(1)}}}$ under the hypothesis \text{\textup{W}[1]}\neq FPT. They also rule out the existence of $(\log n)^{1/{k^{O(1)}}}$-approximation algorithm with running time $f(k)\cdot n^{o(k)}$ for any computable function $f$, assuming ETH, and the existence of $(\log n)^{\frac{1}{{(k+e(\epsilon))^{O(1)}}}}$-approximation algorithms with running time $f(k)\cdot n^{k-\epsilon}$, assuming SETH. Their approach is to first establish a $(\log n)^{\frac{1}{\Omega(k)}}$ gap for MaxCover, then reduce MaxCover to SetCover and obtain a $(\log n)^{\frac{1}{\Omega(k^{2})}}$-gap. This paper presents a new technique which allows us to design simple reductions improving the inapproximation factor to $(1-\epsilon)\cdot\sqrt[k]{\frac{\log n}{\log\log n}}$. The reduction in [8] can get the ratio $(\log n)^{\Omega(1/k)}$ but it has to assume Gap-ETH.

Theorem 1.1**.**

Assuming SETH, for every $\epsilon,\delta\in(0,1)$, sufficiently large $k$111We need large $k$ to get the $\frac{1}{1+\delta}\left(\frac{\log N}{\log\log N}\right)^{\frac{1}{k}}$ gap for small $\delta$. If we want to obtain an $\Theta\left(\sqrt[k]{\frac{\log N}{\log\log N}}\right)$ gap, then our reduction works for all $k\geq 2$. and computable function $f:\mathbb{N}\to\mathbb{N}$, there is no $f(k)\cdot N^{k-\epsilon}$ time algorithm that can, given an $N$-vertex set cover instance $I$, distinguish between

• •

$opt(I)\leq k$,

• •

$opt(I)>\frac{1}{1+\delta}\left(\frac{\log N}{\log\log N}\right)^{\frac{1}{k}}$.

Theorem 1.2**.**

Assuming ETH, there is a constant $\epsilon\in(0,1)$ such that for every $\delta\in(0,1)$ and computable function $f:\mathbb{N}\to\mathbb{N}$, no $f(k)\cdot N^{\epsilon k}$ time algorithm that can, given an $N$-vertex set cover instance $I$ , distinguish between

• •

$opt(I)\leq k$,

• •

$opt(I)>\frac{1}{1+\delta}\cdot\left(\frac{\log N}{\log\log N}\right)^{\frac{1}{k}}$.

Behind these results is a reduction which, given an integer $k$, an $n$-vertex set cover instance $I$ and an integer $h\leq O(\log n/\log\log n)$, produces an $n^{O(1)}\cdot(|U(I)|)^{O(h^{k})}$-vertex instance $I^{\prime}$ in $n^{O(1)}\cdot|U(I)|^{O(h^{k})}$ time such that if $opt(I)\leq k$ then $opt(I^{\prime})\leq k$, otherwise $opt(I^{\prime})>h$. Therefore, to prove the $h$-factor parameterized inapproximability of SetCover, it suffices to show the hardness of SetCover when the input instances have $n^{O(1/h^{k})}$-size universe set. Note that the standard reduction for SETH-hardness of set cover parameterized by the solution size $k$ produces instances $I$ with $|U(I)|=O(k\log|S(I)|)$. With our reduction, this immediately yields the above theorems. Let us not fail to mention that the results of [24] also imply the hardness of SetCover with logarithmic sized universe set assuming the $k$-SUM hypothesis and \text{\textup{W}[1]}\neq FPT hypothesis respectively. Similarly, we can obtain the corresponding inapproximability for set cover based on each of these hypotheses as well. In particular, using a simple trick, we can even rule out $(\log N)^{1/\epsilon(k)}$-approximation FPT-algorithm of set cover for any unbounded computable function $\epsilon$ under \text{\textup{W}[1]}\neq FPT.

Theorem 1.3**.**

Assuming $k$-SUM hypothesis for any $\delta,\epsilon\in(0,1)$, sufficiently large $k$ and computable function $f:\mathbb{N}\to\mathbb{N}$, there is no $f(k)\cdot N^{\lceil k/2\rceil-\epsilon}$ time algorithm that can, given an $N$-vertex set cover instance $I$, distinguish between

• •

$opt(I)\leq k$,

• •

$opt(I)>\frac{1}{1+\delta}\left(\frac{\log N}{\log\log N}\right)^{\frac{1}{k}}$.

Theorem 1.4**.**

Assuming \text{\textup{W}[1]}\neq FPT, for and computable function $f:\mathbb{N}\to\mathbb{N}$ and unbounded computable function $\epsilon:\mathbb{N}\to\mathbb{N}$, there is no $f(k)\cdot N^{O(1)}$-time algorithm that can, given an $N$-vertex set cover instance $I$, distinguish between

• •

$opt(I)\leq k$,

• •

$opt(I)>{\log N}^{1/\epsilon(k)}$.

Technique contribution.

The main technique contribution of this paper is to introduce a gadget that can be used to design gap-producing reductions from the set cover problem to its approximation version and provide a construction of this gadget using $(n,k)$-universal sets. Compared to the reductions in [24], the gap amplification step in this paper is independent of the starting assumptions. This simplifies the proof for showing the inapproximability of the set cover problem. In particular, the inapproximability result in [24] assuming SETH needs some heavy machinery like AG codes to create the gap, while our reduction is completely elementary.

In addition to it simplicity, an important feature of our reduction is that it can be computed by constant depth circuits. Combining this observation with Rossman’s $\Omega(n^{k/4})$ size lower bound for constant depth circuits detecting $k$-clique [33], Wenxin Lai [25] showed that there is no constant-depth circuits of size $f(k)n^{o(\sqrt{k})}$ that can distinguish between a set cover instance with solution size at most $k$ and one whose minimum solution size is at least $({\log n}/{\log\log n})^{1/\binom{k}{2}}$.

Another advantage of our reduction is that it can give hardness approximation result from assumptions that the distributed PCP technique cannot. If we assume that $k$-set-cover with large universe set, say $|U|=n^{1/h(k)^{k}}$, has no $n^{k-\epsilon}$-time algorithm, then our reduction gives $h(k)$ factor hardness of approximation $k$-set-cover in $n^{k-\epsilon}$ time. This cannot be achieved by the distributed PCP technique used in [24] due to known lower bounds in communication complexity of set disjointness.

The gap-gadget we introduce in this paper is similar to the bipartite graphs with threshold property in [26, 27]. Such kind of gadgets may have further applications in proving hardness of approximation for other parameterized problems.

2 Preliminaries

For $n,k\in\mathbb{N}$, an $(n,k)$-universal set is a set of binary strings with length $n$, such that the restriction to any $k$ indices contains all the $2^{k}$ possible binary configurations.

Lemma 2.1**.**

[See Sections 10.5 and 10.6 of [22]] For $k2^{k}\leq\sqrt{n}$, $(n,k)$-universal sets of size $n$ can be computed in $O(n^{3})$ time.

Hypotheses.

Below is a list of hardness hypotheses we will use in this paper.

• •

\text{\textup{W}[1]}\neq FPT: for any computable function $f:\mathbb{N}\to\mathbb{N}$, no algorithm can, given an $n$-vertex graph $G$ and an integer $k$, decide if $G$ contains a $k$-clique in $f(k)\cdot n^{O(1)}$ time.

• •

\text{\textup{W}[2]}\neq FPT: for any computable function $f:\mathbb{N}\to\mathbb{N}$, there is no algorithm which, given an $n$-vertex set cover instance $I$ and an integer $k$, decides if $opt(I)\leq k$ in $f(k)\cdot n^{O(1)}$ time.

• •

Exponential Time Hypothesis (ETH)[19, 20]: there exists a $\delta\in(0,1)$ such that $3$-SAT on $n$ variables cannot be solved in $O(2^{\delta n})$ time.

• •

Strong Exponential Time Hypothesis (SETH)[19, 20] for any $\epsilon\in(0,1)$ there exists $d\geq 3$ such that $d$-SAT on $n$ variables cannot be solved in $O(2^{(1-\epsilon)n})$ time.

• •

$k$-SUM hypothesis ($k$-SUM) [1]: for every $k\geq 2$ and $\epsilon>0$, no $O(n^{\lceil k/2\rceil-\epsilon})$ time algorithm can, given $k$ sets $S_{1},\ldots,S_{k}$ each with $n$ integers in $[-n^{2k},n^{2k}]$, decide if there are $k$ integers $x_{1}\in S_{1},\ldots,x_{k}\in S_{k}$ such that $\sum_{i\in[k]}x_{i}=0$.

We refer the reader to [18, 15] for more information about the parameterized complexity hypotheses. Using the Sparsification lemma [20], we can assume that the instances of $3$-SAT in ETH have $Cn$ clauses for some constant $C$ and the instances of $d$-SAT in SETH have $C_{d,\epsilon}n$ clauses where $C_{d,\epsilon}$ depends on $d$ and $\epsilon$.

3 Reductions

We start with the definition of $(k,n,m,\ell,h)$-gap-gadgets. In Lemma 3.2, we show how to use theses gadgets to create an $(h/k)$-gap for the set cover problem. Lemma 3.5 gives a polynomial time construction of gap-gadgets with $h\leq O({\log n/\log\log n})$ and $\ell=h^{k}$. Since for every input instance $I=(U,S,E)$ of set cover, our reduction runs in time $|S|^{O(1)}|U|^{\ell}$. If $|U|=\Omega(n)$, we can not afford such running time. Our next step is to prove the hardness of set cover with $U=f(k)\cdot(\log n)^{O(1)}$ based on each of the aforementioned hypotheses.

Definition 3.1** ($(k,n,m,\ell,h)$-Gap-Gadget).**

A $(k,n,m,\ell,h)$-Gap-Gadget is a bipartite graph $T=(A,B,E)$ satisfying the following conditions.

(G1)

$A$* is partitioned into $(A_{1},A_{2},\ldots,A_{m})$. For every $i\in[m]$, $|A_{i}|=\ell$.*

(G2)

$B$* is partitioned into $(B_{1},B_{2},\ldots,B_{k})$. For every $j\in[k]$, $|B_{j}|=n$.*

(G3)

For all $b_{1}\in B_{1},b_{2}\in B_{2},\ldots b_{k}\in B_{k}$, there exist $a_{1}\in A_{1},\ldots,a_{m}\in A_{m}$ such that for all $i\in[m]$ and $j\in[k]$, $a_{i}$ is adjacent to $b_{j}$.

(G4)

For all $X\subseteq B$ and $a_{1}\in A_{1},\ldots,a_{m}\in A_{m}$, if every $a_{i}$ has at least $k+1$ neighbors in $X$, then $|X|>h$.

To use this gadget, given a set cover instance $I=(S,U,E)$, we will identify the set $B$ with the set $S$. Then we construct a new set cover instance $I^{\prime}=(S^{\prime},U^{\prime},E^{\prime})$ with $S^{\prime}=S$ such that

• ($\star$)

for any subset $X$ of $S^{\prime}$ that can cover $U^{\prime}$, there must exist a vertex $a_{i}\in A_{i}$ for every $i\in m$ witnessing that $X$ contains a solution of $I$, i.e., there exists $C\subseteq X$ that can cover $U$ in the instance $I$ and all the vertices of $C$ are adjacent to $a_{i}$ in the gap-gadget.

It is easy to check the correctness of this reduction:

If there is a $k$-vertex set $X$ that can cover $U$, then by (G3) we can pick $a_{i}\in A_{i}$ for all $i\in[m]$ such that $a_{i}$ is adjacent to all vertices in $X$. This means that $X$ is also a solution of $I^{\prime}$.

If $opt(I)>k$, then no matter how we pick $a_{i}\in A_{i}$, each $a_{i}$ must have $k+1$ neighbors in $X$. This implies that $X>h$ by (G4).

To achieve ($\star$), we will use the idea of hypercube set system from Feige’s work [17] (which is also used in [24, 8]). For each $i\in[m]$, we construct a set $U^{A_{i}}$. Each element in $U^{A_{i}}$ can be regarded as a function $f:A_{i}\to U$. In the new set cover instance, $f$ is covered by $s\in S$ if there exists $a_{i}\in A_{i}$ such that $a_{i}$ is adjacent to $s$ in the gap-gadget and $f(a_{i})$ is covered by $s$ in $I$. More details can be found in the proof of the following lemma.

Lemma 3.2**.**

There is an algorithm which, given an integer $k$, an instance $I=(S,U,E)$ of SetCover, where $S=S_{1}\cup S_{2}\ldots\cup S_{k}$ and $|S_{i}|=n$ for all $i\in[k]$, and a $(k,n,m,\ell,h)$-Gap-Gadget, outputs a set cover instance $I^{\prime}=(S^{\prime},U^{\prime},E^{\prime})$ with $S^{\prime}=S$ and $U^{\prime}=m|U|^{\ell}$ in $|U|^{\ell}\cdot n^{O(1)}$ time such that

• •

if there exist $s_{1}\in S_{1},\ldots,s_{k}\in S_{k}$ that can cover $U$, then $opt(I^{\prime})\leq k$;

• •

if $opt(I)>k$, then $opt(I^{\prime})>h$.

Proof.

Let $T=(A,B,E_{T})$ be the $(k,n,m,\ell,h)$-Gap-Gadget. Without loss of generality, assume that for all $i\in[k]$ $B_{i}=S_{i}$. The new instance $I^{\prime}=(S^{\prime},U^{\prime},E^{\prime})$ is defined as follows.

• •

$S^{\prime}=S$.

• •

$U^{\prime}=(\bigcup_{i\in[m]}U^{A_{i}})$.

• •

For all $s\in S^{\prime}$ and $f\in U^{A_{i}}$ where $i\in[m]$, $E^{\prime}$ contains $\{s,f\}$ if there exists an $a\in A_{i}$ such that

(E’1)

$\{s,f(a)\}\in E$,

(E’2)

$\{a,s\}\in E_{T}$.

Completeness.

If $opt(I)\leq k$, then there exist $s_{1}\in S_{1},\ldots,s_{k}\in S$ that can cover the whole set $U$. We will show that for every $f\in U^{\prime}$, $f$ is covered by some vertex in $\{s_{1},s_{2},\cdots,s_{k}\}$. Firstly, by (G3), there exist $a_{1}\in A_{1},\ldots,a_{m}\in A_{m}$ such that $a_{i}s_{j}\in E_{T}$ for all $i\in[m]$ and $j\in[k]$. Assume that $f\in U^{A_{i}}$ for some $i\in[m]$. Observe that $f(a_{i})\in U$ must be covered by some $s_{j}$ with $j\in[k]$, i.e., $\{s_{j},f(a_{i})\}\in E$. Since $\{a_{i},s_{j}\}\in E_{T}$ and $\{s_{j},f(a_{i})\}\in E$, according to the definition of $E^{\prime}$, we must have $\{s_{j},f\}\in E^{\prime}$.

Soundness.

Suppose $opt(I)>k$. Let $X\subseteq S^{\prime}$ be a set covering $U^{\prime}$. For every $a\in A$, let $N^{T}(a)$ be the set of neighbors of $a$ in $T$. We have the following claim.

Claim 3.3**.**

For every $i\in[m]$ there exists $a_{i}\in A_{i}$ such that $|N^{T}(a_{i})\cap X|\geq k+1$.

Proof of Claim 3.3. Suppose there exists an $i\in[m]$ such that for all $a\in A_{i}$, $|N^{T}(a)\cap X|\leq k$. Since $opt(I)>k$, every solution of $I$ has size at least $k+1$. It follows that for every $a\in A_{i}$, there exists some $u_{a}\in U$ such that $u_{a}$ is not covered by $N^{T}(a)\cap X$ in the set cover instance $I$. Define a function $f\in U^{A_{i}}$ such that $f(a)=u_{a}$ for every $a\in A_{i}$. We claim that $f$ is not covered by $X$. Otherwise, suppose there exists an $s\in X$ that can cover $f$. According to the definition of $E^{\prime}$, there must exists an $a\in A_{i}$ such that (E’1) and (E’2) hold. However, if $s\in N^{T}(a)\cap X$, then $\{s,f(a)\}=\{s,u_{a}\}\notin E$. On the other hand, if $s\notin N^{T}(a)\cap X$, then $\{a,s\}\notin E_{T}$. In both cases, we obtain contradictions.

$\dashv$

By Claim 3.3, we can pick $a_{i}\in A_{i}$ for each $i\in[m]$ such that every $a_{i}$ has at least $k+1$ neighbors in $X$. By the property of Gap-Gadget, $|X|>h$. ∎

Remark 3.4**.**

Recall that the greedy algorithm can approximate the set cover problem within a $(1+\ln|U|)$-approximation ratio. If one could construct a gap-gadget for parameters satisfying

[TABLE]

then applying the greedy algorithm on input $I^{\prime}$ could decide whether $opt(I)=k$ in $|U|^{\ell}\cdot n^{O(1)}$ time.

It is well known that given a CNF formula $\phi$ on $n$ variables, one can construct a set cover instance $I=(S,U,E)$ with $|U|=O(n)$ and $|S|=\Theta(k2^{n/k})$ in $2^{O(n/k)}$ time such that $\phi$ is satisfiable if and only if $opt(I)=k$. This implies that, assuming ETH there is no algorithm that can construct $(k,|S|,m,\ell,h)$-gap-gadgets with $k(1+\ell ln|U|+\ln m)<h$ and $|U|^{\ell}\leq 2^{o(n)}$ in $2^{o(n)}$ time.

3.1 Construction of Gap-Gadgets

In [27], a similar gadget is used to prove the parameterized complexity of $k$-Biclique. One would wonder if the randomized construction from [27] can be used to construct the gap-gadget in this paper. Informally, the gadget in [27] is a bipartite random graph $T=(A,B,E)$ satisfying the following properties with high probability:

(T1)

a $k$-vertex set in $B$ has $m=n^{\Theta(1/k)}$ common neighbors;

(T2)

any $(k+1)$-vertex set in $B$ has at most $O(k^{2})$ common neighbors.

It is not hard to show that if $Y\subseteq A$ is an $m$-vertex set and every vertex in $Y$ has at least $k+1$ neighbors in $X\subseteq B$, then $|X|\geq\sqrt[k+1]{\frac{|Y|}{O(k^{2})}}$ by (T2) and the pigeonhole principle. We may partition the vertex set $A$ into $m$ parts. Each part contains $n^{1-\Theta(1/k)}$ vertices. This gives us a gap-gadget with large gap $h=\sqrt[k+1]{\frac{m}{O(k^{2})}}$ and $\ell=n^{1-\Theta(1/k)}$. Unfortunately, such gadget does not suit our purpose. We need a gap-gadget with $\ell\leq\log n/\log\log n$. In this section, we provide a construction using universal sets.

Lemma 3.5**.**

There is an algorithm that can, for every $k,h,n\in\mathbb{N}$ with $k\log\log n\leq\log n$ and $h\leq\frac{\log n}{(2+\epsilon)\log\log n}$, compute a $(k,n,n\log h,h^{k},h)$-Gap-Gadget in $O(n^{4})$ time.

Proof.

Let $m=n\log h$ and $K=h\log h$. Note that $(\log m)/2=(\log n+\log\log h)/2\geq(2+\epsilon)h\log h/2\geq\log h+\log\log h+h\log h=\log K+K$, i.e., $K2^{K}\leq\sqrt{m}$. By Lemma 2.1, an $(m,K)$-universal set $S=\{s_{1},s_{2},\ldots,s_{m}\}$ can be constructed in $O(m^{3})\leq O(n^{4})$ time. Partition every $s\in S$ into $n=\frac{m}{\log h}$ blocks so that each block has length $\log h$. Interpret the values of blocks as integers in $[h]$. We obtain an $m\times n$ matrix $M$ by setting the value $M_{r,c}$ equal to the value of the $c$-th block of $s_{r}$. The matrix $M$ satisfies the following conditions.

(M1)

For all $r\in[m]$ and $c\in[n]$, $M_{r,c}\in[h]$.

(M2)

For any set $C\subseteq[n]$ with $|C|\leq h$, there exists a row $r\in[m]$ such that $|\{M_{r,c}:c\in C\}|=|C|$.

Condition (M1) is obvious. To see why (M2) holds, for each $C\subseteq[n]$ with $|C|\leq h$, let $C^{\prime}$ be the set of indices corresponding to the blocks in $C$. Note that $|C^{\prime}|=|C|\log h\leq h\log h=K$. By the property of $(m,K)$-universal set, there exists an $s_{r}\in S$ such that each block in $C$ takes distinct value. It follows that $|\{M_{r,c}:c\in C\}|=|C|$.

For each $i\in[m]$, let

[TABLE]

Note that $|A_{i}|=h^{k}$. For each $j\in[k]$, let $B_{j}=[n]$. Let $T=(A,B,E)$ be a bipartite graph with

• •

$A=\bigcup_{i\in[m]}A_{i}$.

• •

$B=\bigcup_{j\in[k]}B_{j}$.

• •

E=\{\{\vec{a},b\}:\text{\vec{a}\in A_{i},b\in B_{j}$and$M_{i,b}=\vec{a}[j]$for all$j\in[k]}\}.

We will show that $T$ is an $(k,n,m,h^{k},h)$-gap-gadget. Obviously, $T$ satisfies (G1) and (G2).

$T$ satisfies (G3).

For any $b_{1}\in B_{1},b_{2}\in B_{2},\ldots,b_{k}\in B_{k}$. We define $\vec{a}_{i}\in A_{i}$ by setting

[TABLE]

It is routine to check that $\{\vec{a}_{i},b_{j}\}\in E$ for all $i\in[m]$ and $j\in[k]$.

$T$ satisfies (G4).

Let $X\subseteq B$ and $\vec{a}_{1}\in A_{1},\vec{a}_{2}\in A_{2},\ldots,\vec{a}_{m}\in A_{m}$. Suppose for every $i\in[m]$, $\vec{a}_{i}$ has at least $k+1$ neighbors in $X$ and $|X|\leq h$. By (M2), there exists an $r\in[m]$ such that $|\{M_{r,c}:c\in X\}|=|X|$. Since $\vec{a}_{r}$ has at least $k+1$ neighbors in $X$, there exists an $j\in[k]$ such that $\vec{a}_{r}$ has two neighbors $b,b^{\prime}$ in $X\cap B_{j}$. According to the definition of $E$, we must have

[TABLE]

This contradicts the fact that $|\{M_{r,c}:c\in X\}|=|X|$. ∎

The construction above produces gap-gadgets with $\ell=h^{k}$. Note that the parameter $h$ is related to the inapproximation factor we will get for the set cover problem and the running time of our reduction is $n^{O(1)}|U|^{\ell}$. We want to set $h$ as large as possible while keeping the running time of reduction in $f(k)\cdot n^{O(1)}$. Assuming $|U|=g(k)\cdot(\log n)^{O(1)}$, the best we can achieve is $h=(\log n/\log\log n)^{1/k}$.

On the probabilistic construction.

A natural question is, can we construct gap-gadgets with better parameters $h$ and $\ell$, say $\ell=h=o(\log n)$, using the probabilistic method?

Consider the probability space of bipartite random graphs on the vertex sets $A=A_{1}\cup A_{2}\cup\cdots\cup A_{m}$ and $B=B_{1}\cup B_{2}\cup\cdots B_{k}$, where $|A_{i}|=\ell$ and $|B_{j}|=n$. Let $p$ be the edge probability. Each bipartite graph $T$ on $A\cup B$ has probability $Pr[T]=p^{|E(T)|}(1-p)^{|A|\cdot|B|-|E(T)|}$. Fix $k$ vertices $b_{1},b_{2},\ldots,b_{k}$ in $B$. Let $X_{good}$ be the random variable that for every bipartite graph $T$, $X_{good}(T)$ is the number of complete bipartite subgraphs of $T$ which contains exactly one vertex in each $A_{i}$ and the $k$ vertices $b_{1},b_{2},\ldots,b_{k}$ in $B$. Let $X_{bad}$ be the random variable that for every bipartite graph $T$, $X_{bad}(T)$ is the number of subgraphs of $T$ with $h$ vertices in $B$ and one vertex in each $A_{i}$ such that each vertex in $A_{i}$ has at least $k+1$ neighbors in $B$. We want to set the edge probability $p$ so that $\Pr[X_{bad}(T)\geq 1]+\Pr[X_{good}(T)=0]\leq 1-n^{-c}$ for some constant $c>0$. One way to bound $\Pr[X_{bad}(T)\geq 1]$ above is to use Markov’s inequality, which gives us $\Pr[X_{bad}(T)\geq 1]\leq E[X_{bad}]$. So we might assume that $E[X_{bad}]<1$. On the other hand, we have $E[X_{good}]\geq\Pr[X_{good}(T)\geq 1]\geq n^{-c}$. Note that expectations of these two random variables are $E[X_{good}]=\ell^{m}p^{km}$ and $E[X_{bad}]=\ell^{m}\binom{n}{h}p^{(k+1)m}\binom{h}{k+1}^{m}$. We deduce that

[TABLE]

and

[TABLE]

Thus

[TABLE]

We might choose $m$ large enough so that the terms $\frac{c\log n}{mk}$ and $\frac{h\log n}{m(k+1)}$ in (1) become relatively small. In order to make (1) hold, we have to set $\ell\geq h^{O(k^{2})}$. This does not give us better $(k,n,m,\ell,h)$-gap-gadgets.

3.2 Proofs of Theorem 1.1 and Theorem 1.2

Lemma 3.6**.**

There is an algorithm, which given $k\in\mathbb{N}$, $\delta>0$ with $(1+1/k^{3})^{1/k}\leq(1+\delta)/(1+\delta/2)$ and $(1+\delta/2)^{k}\geq 2k^{4}$ and a SAT instance $\phi$ with $n$ variables and $Cn$ clauses, where $n$ is much larger than $k$ and $C$, outputs an integer $N\leq 2^{n/k+n/k^{3}}$ and a set cover instance $I$ satisfying the following conditions in $2^{5n/k}$ time.

• •

$|S(I)|+|U(I)|\leq N$*. *

• •

If $\phi$ is satisfiable, then $opt(I)\leq k$.

• •

If $\phi$ is not satisfiable, then $opt(I)>\frac{1}{1+\delta}\cdot\sqrt[k]{\frac{\log N}{\log\log N}}$.

Proof.

Let $k$ be a positive integer and $\phi$ be a CNF with $n$ variables and $Cn$ clauses. We first construct a set cover instance $I^{\prime}=(S^{\prime},U^{\prime},E^{\prime})$ as follows. Partition the variable set into $k$ parts, each having at most $\lceil n/k\rceil$ variables. For each $i\in[k]$, let $S_{i}$ be the set of assignments to the $i$-th part. Let $S^{\prime}=S_{1}\cup\cdots\cup S_{k}$. Let $U^{\prime}$ be the set consisting of all the clauses of $\phi$ and $k$ additional nodes $u_{1},u_{2},\ldots,u_{k}$. For every $i\in[k]$ and assignment $s\in S_{i}$, we add an edge between $s$ and $u_{i}$. If the assignment $s\in S^{\prime}$ satisfies a clause $u\in U^{\prime}$, we also add an edge between $u$ and $s$. The set cover instance $I^{\prime}$ has the following properties.

• •

If $\phi$ is satisfiable, then $opt(I^{\prime})=k$. Moreover, there exist $k$ vertices $s_{1}\in S_{1},\cdots,s_{k}\in S_{k}$ that can cover the whole set $U^{\prime}$.

• •

If $\phi$ is not satisfiable, then $opt(I^{\prime})>k$.

• •

$|U^{\prime}|=k+Cn$.

• •

$|S^{\prime}|\leq k2^{n/k}$.

Let $M=k2^{n/k}\geq|S|$ and $N=M^{1+1/k^{3}}\leq 2^{n/k+n/k^{3}}$. Note that $\log M/\log\log M\geq n/(k\log n)\geq k$. Applying Lemma 3.5 with $k\leftarrow k$, $n\leftarrow M$, $\ell\leftarrow\frac{\log M}{(1+\delta/2)^{k}\log\log M}$, $h\leftarrow\frac{1}{1+\delta/2}\cdot\sqrt[k]{\frac{\log M}{\log\log M}}$ and $m\leftarrow M\log h\leq M\log\log M$, we obtain a gap-gadget $T$ in $O(M^{4}\leq 2^{5n/k})$ time. Using Lemma 3.2 on $I^{\prime}$ and $T$, we obtain our target set cover instance $I=(S,U,E)$ satisfying the following properties.

• •

If $\phi$ is a yes-instance, then $opt(I)\leq k$.

• •

If $\phi$ is a no-instance, then $opt(I)>\frac{1}{1+\delta/2}\cdot\sqrt[k]{\log M/\log\log M}$. Using $(1+1/k^{3})^{1/k}\leq(1+\delta)/(1+\delta/2)$, we get $opt(I)>\frac{1}{1+\delta}\cdot\sqrt[k]{\log N/\log\log N}$.

• •

$|S|=|S|\leq k2^{n/k}$.

• •

$|U|\leq M\log\log M\cdot|U|^{\frac{\log M}{(1+\delta/2)^{k}\log\log M}}=M\log\log M\cdot(k+Cn)^{\frac{\log M}{(1+\delta/2)^{k}\log\log M}}$.

The number of vertices in $I$ is

[TABLE]

∎

Now we are ready to prove Theorem 1.1. Suppose for some computable function $f$, there is an $f(k)\cdot N^{k-\epsilon}$-time algorithm that can, for every $N$-vertex set cover instance $I$ and every integer $k$, distinguish between $opt(I)\leq k$ and $opt(I)\geq\frac{1}{1+\delta}\cdot\sqrt[k]{\frac{\log N}{\log\log N}}$. For every $\delta\in(0,1)$, choose $k\in\mathbb{N}$ large enough so that $(1+1/k^{3})^{1/k}\leq(1+\delta)/(1+\delta/2)$ and $(1+\delta/2)^{k}\geq 2k^{4}$ hold. Let $\epsilon^{\prime}=1-\epsilon/k+1/k^{2}$, by SETH, there exists an integer $d$ such that $d$-SAT with $n$ variables cannot be solved in $2^{n(1-\epsilon^{\prime})}$-time. Given an instance $\phi$ of $d$-SAT with $n$ variables and $m$ clauses. By the sparsification lemma [20], we can assume that $m=C_{d,\epsilon^{\prime}}\cdot n$ for some constant $C_{d,\epsilon^{\prime}}$ depending on $d$ and $\epsilon^{\prime}$. Without loss of generality, assume that $n$ is much larger than $k$. Applying Lemma 3.6 on $\phi$ and $k$, we obtain a set cover instance $I$ with $N\leq 2^{n/k+n/k^{3}}$ vertices in time $2^{5n/k}\leq 2^{\epsilon n}$ for $k\geq 5/\epsilon$. Then we use the approximation algorithm to decide if $opt(I)\leq k$ or $opt(I)\geq\frac{1}{1+\delta}\cdot\sqrt[k]{\frac{\log N}{\log\log N}}$. Thus we can solve $d$-SAT in time $2^{\epsilon n}+f(k)\cdot N^{k-\epsilon}\leq 2^{\epsilon n}+f(k)\cdot 2^{(n/k+n/k^{3})(k-\epsilon)}\leq 2^{n(1-\epsilon/k+1/k^{2})}=2^{n(1-\epsilon^{\prime})}$, which contradicts SETH.

Theorem 1.2 can be proved similarly. By ETH, there exists $\epsilon>0$ such that $3$-SAT on $n$ variables cannot be solved in $2^{\epsilon n}$ time. Let $\epsilon^{\prime}=\epsilon/2$. For every $3$-SAT instance $\phi$ with $n$ variable and $Cn$ clause, where $n$ is much larger than $k$, apply Lemma 3.6 to obtain a set cover instance $I$ with $N=2^{n/k+n/k^{3}}$ vertices in $2^{5n/k}\leq 2^{\epsilon^{\prime}n}$ time. If there is an $f(k)\cdot N^{\epsilon^{\prime}k}$-time algorithm that can distinguish between $opt(I)\leq k$ and $opt(I)>\frac{1}{1+\delta}\cdot\sqrt[k]{\frac{\log N}{\log\log N}}$, then we can decide whether $\phi$ is satisfiable in time $2^{\epsilon^{\prime}n}+f(k)\cdot 2^{(n/k+n/k^{3})\cdot\epsilon^{\prime}k}\leq 2^{\epsilon n}$.

3.3 Proof of Theorem 1.3

We use a lemma in [2] to reduce $k$-SUM to $k$-VECTOR-SUM over small numbers. Then we present a reduction from $k$-VECTOR-SUM to set cover.

Lemma 3.7** (Lemma 3.1 of [2]).**

Let $k,p,d,s,M\in\mathbb{N}$ satisfy $k<p$, $p^{d}\geq kM+1$, and $s=(k+1)^{d-1}$. There is a collection of mappings $f_{1},\ldots,f_{s}:[0,M]\times[0,kM]\to[-kp,kp]^{d}$, each computable in time $O(poly\log M+k^{d})$, such that for all numbers $x_{1},\ldots,x_{k}\in[0,M]$ and targets $t\in[0,kM]$,

[TABLE]

Lemma 3.8**.**

There is an algorithm which, given $k$ sets $S_{1},S_{2},\ldots,S_{k}$ where $S_{i}$ is a set of $n$ vectors in $[-f(k),f(k)]^{g(k)\log n}$ for some computable functions $f$ and $g$, outputs a set cover instance $I=(S,U,E)$ with $|U|\leq k^{(2f(k))^{k-1}}g(k)\log n$ and $S=S_{1}\cup S_{2}\cup\ldots\cup S_{k}$ in $k^{(2f(k))^{k-1}}g(k)n^{O(1)}$-time such that

• (i)

if there exist $\vec{x}_{1}\in S_{1},\ldots,\vec{x}_{k}\in S_{k}$ such that $\sum_{i\in[k]}\vec{x}_{i}=\vec{0}$, then $\{\vec{x}_{1},\ldots,\vec{x}_{k}\}$ covers $U$;

• (ii)

if the sum of any $k$ vectors $\vec{x}_{1}\in S_{1},\ldots\vec{x}_{k}\in S_{k}$ is not zero, then $opt(I)>k$.

Proof.

Let $D=\{(d_{1},\ldots,d_{k})\in[-f(k),f(k)]^{k}:\sum_{i\in[k]}d_{i}=0\}$. Note that $|D|\leq(2f(k))^{k-1}$. Suppose $D=\{\vec{a}_{1},\ldots,\vec{a}_{|D|}\}$. For every $j\in[g(k)\log n]$, let $U_{j}=[k]^{|D|}$. We define the target set cover instance $I=(S,U,E)$ as follows.

• •

$S=S_{1}\cup\cdots\cup S_{k}$.

• •

$U=\bigcup_{i\in[g(k)\log n]}U_{i}$.

• •

For every $\vec{x}\in S_{i}$ and every $\vec{u}\in U_{j}$, we add an edge $\{\vec{x},\vec{u}\}$ into $E$ if there exists $\ell\in[|D|]$ such that $\vec{u}[\ell]=i$ and $\vec{x}[j]=\vec{a}_{\ell}[i]$.

Completeness.

Suppose there exist $\vec{x}_{1}\in S_{1},\ldots,\vec{x}_{k}\in S_{k}$ such that $\sum_{i\in[k]}\vec{x}_{i}=\vec{0}$. Then for all $j\in[g(k)\log n]$ we have $\vec{x}_{1}[j]+\vec{x}_{2}[j]+\ldots+\vec{x}_{k}[j]=0$, i.e.,

[TABLE]

For all $\vec{u}\in U_{j}$, let $i=\vec{u}[\ell]\in[k]$. Then by (2), $\vec{x}_{i}[j]=\vec{a}_{\ell}[i]$. It follows that $\{\vec{x}_{i},\vec{u}\}\in E$.

Soundness.

Suppose the sum of any $k$ vectors in $S_{1}\cup\cdots\cup S_{k}$ is not zero. Let $X$ be a subset of $S$ with $|X|\leq k$, we need to show that $X$ does not cover $U$. Firstly, we note that if $X\cap S_{i}=\emptyset$ for some $i\in[k]$, then the vector $\vec{u}=(i,i,\ldots,i)\in[k]^{|D|}$ is not covered by any vector in $X$. Now assume that $X=\{\vec{x}_{1},\vec{x}_{2},\ldots,\vec{x}_{k}\}$ and $\vec{x}_{i}\in S_{i}$ for all $i\in[k]$. Since $\sum_{i\in[k]}\vec{x}_{i}\neq\vec{0}$, there exists a $j\in[g(k)\log n]$ such that

[TABLE]

We deduce that

[TABLE]

In other word, for all $\ell\in[|D|]$, there exists an $i_{\ell}\in[k]$ such that

[TABLE]

Define a vector $\vec{u}\in U_{j}$ such that for all $\ell\in[|D|]$,

[TABLE]

Suppose $\vec{u}$ is covered by $x_{i}\in X$, then by the definition, there exists $\ell\in[|D|]$ such that $i=\vec{u}[\ell]=i_{\ell}$ and $\vec{x}_{i_{\ell}}[j]=\vec{a}_{\ell}[i_{\ell}]$, which contradicts (3) and (4). ∎

Proof of Theorem 1.3. Given $k$ sets $S_{1},\ldots,S_{k}$ of integers in $[-n^{2k},n^{2k}]$. Let $p=k^{4k^{c+1}}$, $M=2n^{2k}$ and $d=\log n/k^{c}$. Without loss of generality, assume that $k$ is large and $n$ is much larger than $k$, we have $p^{d}=k^{4k\log n}\geq n^{4k}\geq 2kn^{2k}+1$. On the other hand, for any $\epsilon>0$, we can pick $c$ such that $s=(k+1)^{d}=n^{\log(k+1)/k^{c}}\leq n^{\epsilon/4}$. Applying Lemma 3.7, we obtain a collection of mappings $f_{1},\ldots,f_{s}:[0,M]\times[0,kM]\to[-kp,kp]^{d}$ in $O(poly\log M+k^{d})$ time such that

• •

there exist $x_{1}\in S_{1},\ldots,x_{k}\in S_{k}$ with $\sum_{j\in[k]}x_{j}=0$ if and only if there exist $i\in[s]$ such that $\sum_{j\in[k]}f_{i}(x_{j}+n^{2k},kn^{2k})=\vec{0}$.

Using Lemma 3.5, we construct a $(k,n,O(n\log\log n),\frac{\log n}{(1+\delta/2)^{k}\log\log n},\frac{1}{(1+\delta/2)}\cdot(\frac{\log n}{\log\log n})^{1/k})$-gap-gadget $T$ for some small $\delta>0$. For every $i\in[s]$, and $j\in[k]$, let $S_{j}^{i}=\{f_{i}(x+n^{2k},kn^{2k}):x\in S_{j}\}$. Applying Lemma 3.2 to $S_{1}^{i},S_{2}^{i},\ldots,S_{k}^{i}$ and $T$, we obtain a set cover instance $I_{i}$ with $S(I_{i})=S_{1}^{i}\cup S_{2}^{i}\ldots S_{k}^{i}$ and $|U(I_{i})|\leq n\log\log n\cdot(g(k)\log n)^{\frac{\log n}{(1+\delta/2)^{k}\log\log n}}\leq n^{1+1/k^{3}}$. The set cover instances $I_{1},\ldots,I_{s}$ satisfy the following properties.

• •

If there exist $x_{1}\in S_{1},\ldots,x_{k}\in S_{k}$ with $\sum_{j\in[k]}x_{j}=0$, then there exist $i\in[s]$ and $y_{1}=f_{i}(x_{1}+n^{2k},n^{2k})\in S_{1}^{i}\ldots y_{k}=f_{i}(x_{k}+n^{2k},n^{2k})\in S_{k}^{i}$ such that $y_{1},\ldots,y_{k}$ cover $U(I_{i})$.

• •

If there are no $x_{1}\in S_{1},\ldots,x_{k}\in S_{k}$ with $\sum_{j\in[k]}x_{j}=0$, then for all $i\in[s]$, $opt(I_{i})>\frac{1}{1+\delta/2}\cdot\left(\frac{\log n}{\log\log n}\right)^{1/k}$.

Let $N=n^{1+1/k^{2}}$. We have

[TABLE]

[TABLE]

and

[TABLE]

For every $i\in[s]$, we apply the $f(k)\cdot N^{\lceil k/2\rceil-\epsilon}$-time algorithm to decide if $opt(I_{i})\leq k$ or $opt(I_{i})>\frac{1}{1+\delta}\cdot(\log N/\log\log N)^{1/k}$. If for some $i\in[s]$, it found that $opt(I_{i})\leq k$, then we know that the input instance of $k$-SUM is a yes-instance. The running time is $O(poly\log M+k^{d})+f(k)\cdot N^{\lceil k/2\rceil-\epsilon}\leq O(poly\log M+k^{d})+s\cdot n^{\lceil k/2\rceil-\epsilon+1/k}\leq n^{\lceil k/2\rceil-\epsilon/2}$ for large $k$.

3.4 Proof of Theorem 1.4

Firstly, we give a reduction from Clique to Set-Cover which produces instances with logarithmic sized universe set. The main idea of this reduction is due to Karthik et al. [24].

Lemma 3.9**.**

There is an $n^{O(1)}$-time algorithm which, given an integer $k$, an $n$-vertex graph $G$ with $V(G)=V_{1}\cup V_{2}\cup\cdots\cup V_{k}$ such that $G[V_{i}]$ is an independent set for all $i\in[k]$, outputs a set cover instance $I=(S,U,E)$ with $|U|=k^{O(1)}\log n$ and $S=E(G)=\bigcup_{\{i,j\}\in\binom{[k]}{2}}S_{\{i,j\}}$, where each $S_{\{i,j\}}$ is the set of edges between $V_{i}$ and $V_{j}$, such that

• (i)

if $G$ contains a $k$-clique, then $opt(I)\leq\binom{k}{2}$. Moreover, there exists a $\binom{k}{2}$-sized subset of $S$, which contains exactly one vertex from each $S_{\{i,j\}}$ ($\{i,j\}\in\binom{[k]}{2}$), that can cover $U$;

• (ii)

if $G$ contains no $k$-clique, then $opt(I)>\binom{k}{2}$.

Proof.

We will construct a set cover instance $I$ such that if $G$ has a $k$-clique, then we can select its $\binom{k}{2}$ edges to cover the whole universe set. For every $v\in V(G)$, denote by $encode(v)\in\{0,1\}^{\log n}$ the binary string representation of $v$. For every $\ell\in[\log n]$, the $\ell$th bit of $encode(v)$ is $encode(v)[\ell]$. For every $i\in[k]$, let $\sigma_{i}:[k]\setminus\{i\}\to[k-1]$ be an arbitrary bijection. Our target set cover instance $I=(S,U,E)$ is defined as follows.

• •

$S=E(G)=\bigcup_{\{i,j\}\in\binom{[k]}{2}}S_{\{i,j\}}$, where $S_{\{i,j\}}=\{\{v_{i},v_{j}\}:v_{i}\in V_{i},v_{j}\in V_{j},\{v_{i},v_{j}\}\in E(G)\}$.

• •

$U=[k]\times[k-1]^{\{0,1\}}\times[\log n]$.

• •

For $s=\{v_{i},v_{j}\}\in S$ and $u=(i,f,\ell)\in U$ we add $\{s,u\}$ into $E$ if

[TABLE]

The set cover instance $I$ satisfies the following conditions.

• •

If $G$ contains a $k$-clique, then there exists a $\binom{k}{2}$-sized subset of $S$ which contains exactly one vertex from each $S_{\{i,j\}}$ ($\{i,j\}\in\binom{[k]}{2}$) that can cover $U$. Suppose that $v_{1}\in V_{1},\ldots,v_{k}\in V_{k}$ induce a $k$-clique. Let $X=\{\{v_{i},v_{j}\}:\{i,j\}\in\binom{[k]}{2}\}$. We will show that $X$ covers the whole set $U$. For any $(i,f,\ell)\in U$, let $b=encode(v_{i})[\ell]$. Since $f(b)\in[k-1]$, there must exist a $j\in[k]\setminus\{i\}$ such that $\sigma_{i}(j)=f(b)$. By the definition of $E$, $\{v_{i},v_{j}\}$ is adjacent to $(i,f,\ell)$ .

• •

If $G$ does not contain a $k$-clique, then $opt(I)>\binom{k}{2}$. Let $X\subseteq S$ be a set such that $|X|\leq\binom{k}{2}$ and $X$ covers $U$.

For each $\{i,j\}\in\binom{[k]}{2}$, define

[TABLE]

We claim that for every $\{i,j\}\in\binom{[k]}{2}$, $|X_{\{i,j\}}|>0$. Otherwise let $f(0)=f(1)=\sigma_{i}(j)$ and consider the vertex $(i,f,1)\in U$. According to the definition of $E$, if a vertex $\{v,u\}\in S$ covers $(i,f,1)$, then either $v$ or $u$ must be in $V_{i}$. Let us assume $v\in V_{i}$ and $u\in V_{j^{\prime}}$ for some $j^{\prime}\in[k]\setminus\{i\}$. We must have $f(encode(v_{i})[1])=\sigma_{i}(j^{\prime})$. However, if $j\neq j^{\prime}$, then $f(0)=f(1)=\sigma_{i}(j)\neq\sigma_{i}(j^{\prime})$.

Since $\binom{k}{2}\geq|X|=\sum_{\{i,j\}\in\binom{[k]}{2}}|X_{\{i,j\}}|$ and $|X_{\{i,j\}}|>0$, we conclude that $|X_{\{i,j\}}|=1$ for all $\{i,j\}\in\binom{[k]}{2}$.

For every $i\in[k]$ and distinct $j,j^{\prime}\in[k]\setminus\{i\}$, let $\{\{v,v_{j}\}\}=X_{\{i,j\}}$ and $\{\{v^{\prime},v_{j^{\prime}}\}\}=X_{i,j^{\prime}}$, where $v,v^{\prime}\in V_{i}$, we claim that $v=v^{\prime}$. Otherwise, since $v\neq v^{\prime}$ there exists $\ell\in[\log n]$ such that $encode(v)[\ell]\neq encode(v^{\prime})[\ell]$. Now consider a function $f$ with $f(encode(v^{\prime})[\ell])=\sigma_{i}(j)$ and $f(encode(v)[\ell])=\sigma_{i}(j^{\prime})$. The vertex $(i,f,\ell)$ must be covered by some $\{x,y\}$ with $x\in V_{i}$ and $y\in V_{h}$ such that $\sigma_{i}(h)=f(encode(v)[\ell])\in\{\sigma_{i}(j),\sigma_{i}(j^{\prime})\}$. We must have $y\in V_{j}$ or $y\in V_{j^{\prime}}$. Since $|X_{\{i,j\}}|=|X_{\{i,j^{\prime}\}}|=1$, we deduce that either $\{x,y\}=\{v,v_{j}\}$ or $\{x,y\}=\{v^{\prime},v_{j^{\prime}}\}$. However, if $\{x,y\}=\{v,v_{j}\}$, we must have $\sigma_{i}(j)=f(encode(v)[\ell])=\sigma_{i}(j^{\prime})\neq\sigma_{i}(j)$, a contradiction. Similarly, if $\{x,y\}=\{v^{\prime},v_{j^{\prime}}\}$, then $\sigma_{i}(j^{\prime})=f(encode(v^{\prime})[\ell])=\sigma_{i}(j)\neq\sigma_{i}(j^{\prime})$. We conclude that the vertex $(i,f,\ell)$ can not be covered by $X$.

Now we have for every $i\in[k]$, there exists a $v_{i}\in V_{i}$ such that

[TABLE]

Obviously, for every $\{i,j\}\in\binom{[k]}{2}$, $\{\{v_{i},v_{j}\}\}=X_{\{i,j\}}$. This implies that $\{v_{1},v_{2},\ldots,v_{k}\}$ is a $k$-clique in $G$.

∎

Proof of Theorem 1.4. Given an $n$-vertex graph $G$ and a positive integer $k$, we invoke Lemma 3.9 to obtain a set cover instance $I=(S,U,E)$ with $|S|=|E(G)|$ and $|U|\leq k^{3}\log n$ satisfying (i) and (ii). Let $m=|S|$. Then we use Lemma 3.5 to construct a $(\binom{k}{2},m,n^{O(1)},\frac{\log m}{\log\log m},\frac{\log m}{\log\log m}^{1/\binom{k}{2}})$-gap-gadget $T$ in $m^{O(1)}=n^{O(1)}$ time. Applying Lemma 3.2 on $I$ and $T$, we finally obtain our target set cover instance $I^{\prime}=(S^{\prime},U^{\prime},E^{\prime})$ with the following properties:

• •

if $G$ has a $k$-clique, then $opt(I^{\prime})=\binom{k}{2}$,

• •

if $G$ has no $k$-clique, then $opt(I^{\prime})>\left(\frac{\log m}{\log\log m}\right)^{1/\binom{k}{2}}$,

• •

$|S^{\prime}|=|E(G)|=m$,

• •

$|U^{\prime}|=(k^{3}\log n)^{\log m/\log\log m}=m^{1+o(1)}$.

Let $N=|U^{\prime}|+|S^{\prime}|$. We have $N=n^{O(1)}$. Since $\epsilon$ is an unbounded computable function, there is a computable function $g:\mathbb{N}\to\mathbb{N}$ such that $k^{\prime}=g(k)>\binom{k}{2}$ and $\epsilon(k^{\prime})>\binom{k}{2}$. When $n$ is large enough,

[TABLE]

Any $f(k^{\prime})\cdot N^{O(1)}$ time algorithm that can distinguish between $opt(I^{\prime})\leq k^{\prime}$ and $opt(I^{\prime})>(\log N)^{\frac{1}{\epsilon(k^{\prime})}}$ can be used to decide if an input graph $G$ has $k$-clique in $f(g(k))n^{O(1)}$ time.

4 Conclusion

We have improved the hardness approximation factor for the parameterized set cover problem using a simple reduction. Our result shows that in order to prove inapproximability of parameterized set cover, it suffices to prove the hardness of set cover problem with small universe set. A natural question is:

Is there any algorithm that can, given an $n$-vertex set cover instance $I$ and an integer $k$, outputs a new instance $I^{\prime}$ and an integer $k^{\prime}$ in $f(k)\cdot n^{O(1)}$ time for some computable function $f:\mathbb{N}\to\mathbb{N}$ such that

• •

$k^{\prime}=g(k)$ for some computable function $g:\mathbb{N}\to\mathbb{N}$,

• •

$opt(I)\leq k$ if and only if $opt(I^{\prime})\leq k^{\prime}$,

• •

$|U(I^{\prime})|\leq h(k)\cdot(\log|S(I^{\prime})|)^{O(1)}$ for some computable function $h:\mathbb{N}\to\mathbb{N}$.

A positive answer to the above question would imply that SetCover parameterized by the optimum solution size has no $(\log n)^{1/\epsilon(k)}$-approximation FPT algorithm assuming \text{\textup{W}[2]}\neq FPT. Of course, if we just want a $\rho$-factor hardness of approximation, then it suffices to have $|U(I^{\prime})|\leq h(k)|S(I^{\prime})|^{O(1/\rho^{k})}$. Note that using Dynamic Programming, SetCover can be solved in $2^{|U(I)|}(|U(I)|+|S(I)|)^{O(1)}$ time [12]. We do not expect to reduce the size of universe set below $o(k\log n)$ under ETH.

Our hardness result is far from matching the $(1+\ln n)$ approximation ratio of the greedy algorithm in polynomial time. Could it be the case that there exists a $(\ln n)^{1/\rho(k)}$-approximation algorithm for SetCover with running time $n^{k-\epsilon}$? What is the best approximation ratio we can achieve for parameterized set cover in $n^{k-\epsilon}$ time?

Acknowledgement This work was supported by JSPS KAKENHI Grant Number JP18H05291. The author wishes to thank the anonymous referees for their detailed comments.