Multistage Vertex Cover

TL;DR

This paper introduces Multistage Vertex Cover, a temporal graph problem requiring small vertex covers across layers with limited differences, revealing its computational hardness and fixed-parameter tractability results.

Contribution

It initiates the study of Multistage Vertex Cover, analyzing its complexity and identifying fixed-parameter tractability in certain cases.

Findings

Multistage Vertex Cover is NP-hard even in restricted settings.

Certain parameterizations allow fixed-parameter tractability.

The problem differs from classic and other dynamic vertex cover variants.

Abstract

Covering all edges of a graph by a small number of vertices, this is the NP-complete Vertex Cover problem. It is among the most fundamental graph-algorithmic problems. Following a recent trend in studying temporal graphs (a sequence of graphs, so-called layers, over the same vertex set but, over time, changing edge sets), we initiate the study of Multistage Vertex Cover. Herein, given a temporal graph, the goal is to find for each layer of the temporal graph a small vertex cover and to guarantee that two vertex cover sets of every two consecutive layers differ not too much (specified by a given parameter). We show that, different from classic Vertex Cover and some other dynamic or temporal variants of it, Multistage Vertex Cover is computationally hard even in fairly restricted settings. On the positive side, however, we also spot several fixed-parameter tractability results based on some of the most natural parameterizations.

Tables

Table 1. Table 1 : Overview of our results. The column headings describe the restrictions on the input and each row corresponds to a parameter. p-NP-hard, PK, and NoPK abbreviate para-NP-hard, polynomial-size problem kernel, and no problem kernel of polynomial size unless  coNP ⊆ NP / poly coNP NP poly \mathrm{coNP}\subseteq\mathrm{NP}/\operatorname{poly} .

	general layers		tree layers	one-edge layers
		$\ell \ge 2k$
	$\mathrm{NP}$-hard	$\mathrm{NP}$-hard	$\mathrm{NP}$-hard	$\mathrm{NP}$-hard
			(Thm. 4.1(i))	(Thm. 4.1(ii))
$\tau$	p-$\mathrm{NP}$-hard	p-$\mathrm{NP}$-hard	p-$\mathrm{NP}$-hard	FPT, PK
	(Thm. 4.1)	(Thm. 4.1)	(Thm. 4.1)	(LABEL:thm:preproctau)
$k$	$\mathrm{XP}$, $W\left[1\right]$-h.,	FPT†, NoPK	$\mathrm{XP}$, $W\left[1\right]$-h.	open, NoPK
	(LABEL:thm:xpwhardness)	(Section 3, LABEL:thm:preprock)	(LABEL:thm:xpwhardness, LABEL:rem:whardnesstree)	(LABEL:thm:preprock)
$k+\tau$	FPT, PK	FPT, PK	FPT, PK	FPT, PK
	(LABEL:thm:PKktau)	(LABEL:thm:PKktau)	(LABEL:thm:PKktau)	(LABEL:thm:PKktau)

Full text

1 Introduction

Vertex Cover asks, given an undirected graph $G$ and an integer $k\geq 0$, whether at most $k$ vertices can be deleted from $G$ such that the remaining graph contains no edge. Vertex Cover is NP-complete and it is a formative problem of algorithmics and combinatorial optimization. We study a time-dependent, “multistage” version, namely a variant of Vertex Cover on temporal graphs. A temporal graph $\mathcal{G}$ is a tuple $(V,\mathcal{E},\tau)$ consisting of a set $V$ of vertices, a discrete time-horizon $\tau$, and a set of temporal edges $\mathcal{E}\subseteq\binom{V}{2}\times\{1,\ldots,\tau\}$. Equivalently, a temporal graph $\mathcal{G}$ can be seen as a vector $(G_{1},\ldots,G_{\tau})$ of static graphs (layers), where each graph is defined over the same vertex set $V$. Then, our specific goal is to find a small vertex cover $S_{i}$ for each layer $G_{i}$ such that the size of the symmetric difference $S_{i}\triangle S_{i+1}=(S_{i}\setminus S_{i+1})\cup(S_{i+1}\setminus S_{i})$ of the vertex covers $S_{i}$ and $S_{i+1}$ of every two consecutive layers $G_{i}$ and $G_{i+1}$ is small. Formally, we thus introduce and study the following problem (see Figure 1 for an illustrative example).

Multistage Vertex Cover

Input:

A temporal graph $\mathcal{G}=(V,\mathcal{E},\tau)$ and two integers $k\in\mathbb{N},\ell\in\mathbb{N}_{0}$.

Question:

Is there a sequence $\mathcal{S}=(S_{1},\ldots,S_{\tau})$ such that

1. (i)

for all $i\in\{1,\ldots,\tau\}$, it holds true that $S_{i}\subseteq V$ is a size-at-most-$k$ vertex cover for layer $G_{i}$, and 2. (ii)

for all $i\in\{1,\ldots,\tau-1\}$, it holds true that $|S_{i}\triangle S_{i+1}|\leq\ell$?

Throughout this paper we assume that $0<k<|V|$ because otherwise we have a trivial instance. In our model, we follow the recently proposed multistage [BampisELP18, GuptaTW14, BampisET19, EMS14, BEK19, HHKNRS19, FNSZ20, CTW20] view on classical optimization problems on temporal graphs.

In general, the motivation behind a multistage variant of a classical problem such as Vertex cover is that the environment changes over time (here reflected by the changing edge sets in the temporal graph) and a corresponding adaptation of the current solution comes with a cost. In this spirit, the parameter $\ell$ in the definition of Multistage Vertex Cover allows to model that only moderate changes concerning the solution vertex set may be wanted when moving from one layer to the subsequent one. Indeed, in this sense $\ell$ can be interpreted as a parameter measuring the degree of (non-)conservation [HN13, AEFRS15].

It is immediate that Multistage Vertex Cover is $\mathrm{NP}$-hard as it generalizes Vertex Cover ($\tau=1$). We will study its parameterized complexity regarding the problem-specific parameters $k$, $\tau$, $\ell$, and some of their combinations, as well as restrictions to temporal graph classes [CasteigtsFQS12, FluschnikMNZ18].

Related Work.

The literature on vertex covering is extremely rich, even when focusing on parameterized complexity studies. Indeed, Vertex Cover can be seen as “drosophila” of parameterized algorithmics. Thus, we only consider Vertex Cover studies closely related to our setting. First, we mention in passing that Vertex Cover is studied in dynamic graphs [IwataO14, AlmanMW17] and graph stream models [ChitnisCEHMMV16]. More importantly for our work, Akrida et al. [AkridaMSZ18] studied a variant of Vertex Cover on temporal graphs. Their model significantly differs from ours: they want an edge to be covered at least once over every time window of some given size $\Delta$. That is, they define a temporal vertex cover as a set $S\subseteq V\times\{1,\ldots,\tau\}$ such that, for every time window of size $\Delta$ and for each edge $e=\{v,w\}$ appearing in a layer contained in the time window, it holds that $(v,t)\in S$ or $(w,t)\in\leavevmode\nobreak\ S$ for some $t$ in the time window with $(e,t)\in\mathcal{E}$. For their model, Akrida et al. ask whether such an $S$ of small cardinality exists. Note that if $\Delta>1$, then for some $t\in\{1,\ldots,\tau\}$ the set $S_{t}\coloneqq\{v\mid(v,t)\in S\}$ is not necessarily a vertex cover of layer $G_{t}$. For $\Delta=1$, each $S_{t}$ must be a vertex cover of $G_{t}$. However, in Akrida et al.’s model the size of each $S_{t}$ as well as the size of the symmetric difference between each $S_{t}$ and $S_{t+1}$ may strongly vary. They provide several hardness results and algorithms (mostly referring to approximation or exact algorithms, but not to parameterized complexity studies).

A second related line of research, not directly referring to temporal graphs though, studies reconfiguration problems which arise when we wish to find a step-by-step transformation between two feasible solutions of a problem such that all intermediate results are feasible solutions as well [ito2011complexity, gopalan2009connectivity]. Among other reconfiguration problems, Mouawad et al. [mouawad2017parameterized, mouawad2018vertex] studied Vertex Cover Reconfiguration: given a graph $G$, two vertex covers $S$ and $T$ each of size at most $k$, and an integer $\tau$, the question is whether there is a sequence $(S=S_{1},\dots,S_{\tau}=T)$ such that each $S_{t}$ is a vertex cover of size at most $k$. The essential difference to our model is that from one “sequence element” to the next only one vertex may be changed and that the input graph does not change over time. Indeed, there is an easy reduction of this model to ours while the opposite direction is unlikely to hold. This is substantiated by the fact that Mouawad et al. [mouawad2017parameterized] showed that Vertex Cover Reconfiguration is fixed-parameter tractable when parameterized by vertex cover size $k$ while we show W[1]-hardness for the corresponding case of Multistage Vertex Cover.

Finally, there is also a close relation to the research on dynamic parameterized problems [AEFRS15, KST18]. Krithika et al. [KST18] studied Dynamic Vertex Cover where one is given two graphs on the same vertex set and a vertex cover for one of them together with the guarantee that the cardinality of the symmetric difference between the two edge sets is upper-bounded by a parameter $d$. The task then is to find a vertex cover for the second graph that is “close enough” (measured by a second parameter) to the vertex cover of the first graph. They show fixed-parameter tractability and a linear kernel with respect to $d$.

Our Contributions.

Our results, focusing on the three perhaps most natural parameters, are summarized in Table 1.

We highlight a few specific results. Multistage Vertex Cover remains $\mathrm{NP}$-hard even if every layer consists of only one edge; not surprisingly, the corresponding hardness reduction exploits an unbounded number $\tau$ of time layers. If one only has two layers, however, one of them being a tree and the other being a path, then again Multistage Vertex Cover already becomes NP-hard. Multistage Vertex Cover parameterized by solution size $k$ is fixed-parameter tractable if $\ell\geq 2k$, but becomes $\mathrm{W}[1]$-hard if $\ell<2k$. Considering the tractability results for Dynamic Vertex Cover [KST18] and Vertex Cover Reconfiguration [mouawad2017parameterized], this hardness is surprising, and it is our most technical result. Furthermore, Multistage Vertex Cover parameterized by $k$ with $\ell\geq 2k$ does not admit a problem kernel of polynomial size unless $\mathrm{coNP}\subseteq\mathrm{NP}/\operatorname{poly}$. Finally, for the combined parameter $k+\tau$ we obtain polynomial-sized problem kernels (and thus fixed-parameter tractability) in all cases without any further constraints.

Outline.

In Section 2, we provide some preliminaries. For Multistage Vertex Cover, we give some first and general observations in Section 3, study the parameterized complexity regarding $k$ in LABEL:sec:paramvc, and discuss the possibilities for efficient data reduction in LABEL:sec:dataredu. We conclude in LABEL:sec:conclusion.

2 Preliminaries

We denote by $\mathbb{N}$ and $\mathbb{N}_{0}$ the natural numbers excluding and including zero, respectively. For two sets $A$ and $B$, we denote by $A\triangle B\coloneqq(A\setminus B)\cup(B\setminus A)=(A\cup B)\setminus(A\cap B)$ the symmetric difference of $A$ and $B$, and by $A\uplus B$ the disjoint union of $A$ and $B$.

Temporal Graphs.

A temporal graph $\mathcal{G}$ is a tuple $(V,\mathcal{E},\tau)$ consisting of the set $V$ of vertices, the set $\mathcal{E}$ of temporal edges, and a discrete time-horizon $\tau$. A temporal edge $e$ is an element in $\binom{V}{2}\times\{1,\ldots,\tau\}$. Equivalently, a temporal graph $\mathcal{G}$ can be defined as a vector of static graphs $(G_{1},\ldots,G_{\tau})$, where each graph is defined over the same vertex set $V$. We also denote by $V(\mathcal{G})$, $\mathcal{E}(\mathcal{G})$, and $\tau(\mathcal{G})$ the set of vertices, the set of temporal edges, and the discrete time-horizon of $\mathcal{G}$, respectively. The underlying graph $G_{\downarrow}=G_{\downarrow}(\mathcal{G})$ of a temporal graph $\mathcal{G}$ is the static graph with vertex set $V(\mathcal{G})$ and edge set $\{e\mid\exists t\in\{1,\ldots,\tau(\mathcal{G})\}:(e,t)\in\mathcal{E}(\mathcal{G})\}$.

Parameterized Complexity Theory.

Let $\Sigma$ be a finite alphabet. A parameterized problem $L$ is a subset $L\subseteq\{(x,k)\in\Sigma^{*}\times\mathbb{N}_{0}\}$. An instance $(x,k)\in\Sigma^{*}\times\mathbb{N}_{0}$ is a yes-instance of $L$ if and only if $(x,k)\in L$ (otherwise, it is a no-instance). Two instances $(x,k)$ and $(x^{\prime},k^{\prime})$ of parameterized problems $L,L^{\prime}$ are equivalent if $(x,k)\in L\iff(x^{\prime},k^{\prime})\in L^{\prime}$. A parameterized problem $L$ is fixed-parameter tractable (FPT) if for every input $(x,k)$ one can decide whether $(x,k)\in L$ in $f(k)\cdot|x|^{O(1)}$ time, where $f$ is some computable function only depending on $k$. A parameterized problem $L$ is in $\mathrm{XP}$ if for every instance $(x,k)$ one can decide whether $(x,k)\in L$ in time $|x|^{f(k)}$ for some computable function $f$ only depending on $k$. A $\mathrm{W}[1]$-hard parameterized problem is fixed-parameter intractable unless $\mathrm{FPT}$=$\mathrm{W}[1]$.

Given a parameterized problem $L$, a kernelization is an algorithm that maps any instance $(x,k)$ of $L$ in time polynomial in $|x|+k$ to an instance $(x^{\prime},k^{\prime})$ of $L$ (the problem kernel) such that

(i) $(x,k)\in L\iff(x^{\prime},k^{\prime})\in L$, and

(ii) $|x^{\prime}|+k^{\prime}\leq f(k)$for some computable function $f$ (the size of the problem kernel) only depending on $k$.

We refer to Downey and Fellows [downey2013fundamentals] and Cygan et al. [cygan2015parameterized] for more material on parameterized complexity.

3 Basic Observations

In this section, we state some preliminary simple-but-useful observations on Multistage Vertex Cover and its relation to Vertex Cover.

Observation \thetheorem.

Every instance $(\mathcal{G},k,\ell)$ of Multistage Vertex Cover with $k\geq\sum_{i=1}^{\tau(\mathcal{G})}|E(G_{i})|$ is a yes-instance.

Proof.

It is easy to see that a graph with $m$ edges always admits a vertex cover of size $m$. Hence, there is a vertex cover $S\subseteq V$ of size $k$ of $G_{\downarrow}(\mathcal{G})$, and hence, $S$ is a vertex cover for each layer. The vector $(S_{1},\ldots,S_{\tau})$ with $S_{i}=S$ for all $i\in\{1,\ldots,\tau\}$ is a solution for every $\ell\geq 0$. ∎

Next, we state that if we are facing a yes-instance, then we can assume that there exists a solution where each layer’s vertex cover is either of size $k$ or $k-1$.

Observation \thetheorem.

Let $(\mathcal{G},k,\ell)$ be an instance of Multistage Vertex Cover. If $(\mathcal{G},k,\ell)$ is a yes-instance, then there is a solution $\mathcal{S}=(S_{1},\ldots,S_{\tau})$ such that $|S_{1}|=k$ and $k-1\leq|S_{i}|\leq k$ for all $i\in\{1,\ldots,\tau\}$.

Proof.

We first show that there is a solution $\mathcal{S}=(S_{1},\ldots,S_{\tau})$ for $I\coloneqq(\mathcal{G},k,\ell)$ such that $|S_{1}|=k$. Towards a contradiction assume that such a solution does not exist. Let $\mathcal{S}=(S_{1},\ldots,S_{\tau})$ be a solution such that $|S_{1}|$ is maximal over all solutions for $I$. Let $i\in\{1,\ldots,\tau\}$ be the maximum index such that $S_{j}\subseteq S_{j-1}$, for all $j\in\{2,\ldots,i\}$. If $i=\tau$, then we have that $|S_{j}|\leq|S_{1}|<k$ for all $j\in\{1,\ldots,\tau\}$. Hence, we can find a subset $X\subseteq V\setminus S_{1}$ such that $(S_{1}\cup X,\ldots,S_{\tau}\cup X)$ is a solution. This contradicts $|S_{1}|$ being maximal. Now let $i<\tau$. Hence, there is a vertex $v\in S_{i+1}\setminus S_{i}$. Now we can adjust the solution by adding $v$ to $S_{j}$ for all $j\in\{1,\dots,i\}$. This contradicts $|S_{1}|$ being maximal. Hence, there is a solution $\mathcal{S}=(S_{1},\ldots,S_{\tau})$ such that $|S_{1}|=k$.

Let $\Psi$ be the set of solutions such that the first vertex cover is of size $k$. Assume towards a contradiction that all solutions in $\Psi$ contain a vertex cover smaller than $k-1$. Let $\Psi_{i}\subseteq\Psi$ be the set of solutions such that for each $(S_{1},\ldots,S_{\tau})\in\Psi_{i}$ we have that $|S_{i}|<k-1$ and $|S_{j}|\geq k-1$ for all $j\in\{1,\ldots,i-1\}$. Let $i\in\{1,\ldots,\tau\}$ be maximal such that $\Psi_{i}\not=\emptyset$. Furthermore, let $\mathcal{S}=(S_{1},\ldots,S_{\tau})\in\Psi_{i}$ such that $|S_{i}|$ is maximal over all solutions in $\Psi_{i}$. Hence, there is a vertex $v\in S_{i-1}\setminus S_{i}$. We distinguish two cases.

(a):

Assume that there is a $p\in\{i+1,\dots,\tau\}$ such that there is a $w\in S_{p}\setminus S_{p-1}$ and $S_{j}\subseteq S_{j-1}$ for all $j\in\{i+1,\ldots,p-1\}$. The idea now is to keep $v$ and add $w$ in the $i$-th layer and then remove $v$ in the $p$-th layer. We can achieve this by simply setting $S_{q}\coloneqq S_{q}\cup\{v,w\}$ for all $q\in\{i,\ldots,p-1\}$. Note that this is a solution which either contradicts that $|S_{i}|$ is maximal or that $i$ is maximal.

(b):

Now assume that $S_{j}\subseteq S_{j-1}$ for all $j\in\{i+1,\ldots,\tau\}$. In this case we take an arbitrary vertex $w\in V\setminus S_{i}$ and set $S_{q}\coloneqq\{v,w\}$ for all $q\in\{i,\ldots,\tau\}$. This contradicts $i$ being maximal.

∎

With the next two observations, we show that the special case of Multistage Vertex Cover where $\ell=0$ is equivalent to Vertex Cover under polynomial-time many-one reductions.

Observation \thetheorem.

There is a polynomial-time algorithm that maps any instance $(G=(V,E),k)$ of Vertex Cover to an equivalent instance $(\mathcal{G},k,\ell)$ of Multistage Vertex Cover where $\ell=0$ and every layer $G_{i}$ contains only one edge.

Proof.

Let the edges $E=\{e_{1},\ldots,e_{m}\}$ of $G$ be ordered in an arbitrary way. Set $\tau=m$ and $\ell=0$. Set $G_{i}=(V,\{e_{i}\})$ for each $i\in\{1,\ldots,\tau\}$. We claim that $(G=(V,E),k)$ is a yes-instance of Vertex Cover if and only if $(\mathcal{G},k,\ell)$ is a yes-instance of Multistage Vertex Cover.

$(\Rightarrow)$ Let $S$ be a vertex cover of $G$ of size at most $k$. Set $S_{i}\coloneqq S$ for all $i\in\{1,\ldots,\tau\}$. Clearly, $S_{i}$ is a vertex cover of $G_{i}$ for all $i\in\{1,\ldots,\tau\}$ of size at most $k$. Moreover, by construction, $|S_{i}\triangle S_{i+1}|\leq 1$ for all $i\in\{1,\ldots,\tau-1\}$. Hence, $(S_{1},\ldots,S_{\tau})$ forms a solution to $(\mathcal{G},k,\ell)$.

$(\Leftarrow)$ Let $\mathcal{S}=(S_{1},\ldots,S_{\tau})$ be a solution to $(\mathcal{G},k,\ell)$. Observe that $|\bigcup_{i}S_{i}|\leq k$. It follows that there are at most $k$ vertices covering all edges of the layers $G_{i}$, that is, $E=\bigcup_{i=1}^{\tau}E(G_{i})$, and hence they cover all edges of $G$. ∎

Observation \thetheorem.

There is a polynomial-time algorithm that maps any instance $(\mathcal{G},k,\ell)$ of Multistage Vertex Cover with $\ell=0$ to an equivalent instance $(G,k)$ of Vertex Cover.

Proof.

Now let $(\mathcal{G}=(V,\mathcal{E},\tau),k,0)$ be an arbitrary instance of Multistage Vertex Cover. Construct the instance $(G_{\downarrow},k)$ of Vertex Cover. We claim that $(\mathcal{G},k,0)$ is a yes-instance if and only if $(G_{\downarrow},k)$ is a yes-instance.

$(\Leftarrow)$ Let $S\subseteq V$ be a vertex cover of size at most $k$. Since $S$ is a vertex cover for $G_{\downarrow}$, $S$ covers each layer of $\mathcal{G}$. Hence, $S_{i}\coloneqq S$ for all $i\in\{1,\ldots,\tau\}$ forms a solution to $(\mathcal{G},k,0)$.

$(\Rightarrow)$ Let $(S_{1},\ldots,S_{\tau})$ be a solution to $(\mathcal{G},k,0)$. Clearly, since $\ell=0$, we have that $S_{i}=S_{j}$ for all $i,j\in\{1,\ldots,\tau\}$. It is not difficult to see that $S\coloneqq S_{1}$ is a vertex cover for $G_{\downarrow}$, and hence the claim follows. ∎

Finally, the special case of Multistage Vertex Cover with $\ell\geq 2k$ (that is, where vertex covers of any two consecutive layers can be even disjoint) is Turing-reducible to Vertex Cover.

Observation \thetheorem.

Any instance $(\mathcal{G},k,\ell)$ of Multistage Vertex Cover with $\ell\geq 2k$ and $\mathcal{G}=(G_{1},\ldots,G_{\tau})$ can be decided by deciding each instance of the set $\{(G_{i},k)\mid 1\leq i\leq\tau\}$ of Vertex Cover-instances.

Proof.

For each of the layers $G_{i}$, $i\in\{1,\ldots,\tau\}$, we can construct an instance of Vertex Cover of the form $(G_{i},k)$. We can solve each instance independently, since the symmetric difference of any two size-at-most-$k$ solutions is at most $2k\leq\ell$. ∎

4 Hardness for Restricted Input Instances

Multistage Vertex Cover is $\mathrm{NP}$-hard as it generalizes Vertex Cover ($\tau=1$). In this section we prove that Multistage Vertex Cover remains $\mathrm{NP}$-hard on inputs with only two layers, one consisting of a path and the other consisting of a tree, and on inputs where every layer consists only of one edge.

Theorem 4.1**.**

Remark 4.2**.**

Theorem 4.1(i) is tight regarding $\tau$ since

Problem 4**.**

Vertex Cover remains $\mathrm{NP}$-complete on cubic Hamiltonian graphs when a Hamiltonian cycle is additionally given in the input [FleischnerSS10]:111A graph is cubic if each vertex is of degree exactly three; A graph is Hamiltonian if it contains a subgraph being a Hamiltonian cycle, that is, a cycle that visits each vertex in the graph exactly once.

Hamiltonian Cubic Vertex Cover (HCVC)

Input:

An undirected, cubic, Hamiltonian graph $G=(V,E)$, an integer $k\in\mathbb{N}$, and a Hamiltonian Cycle $C=(V,E^{\prime})$ of $G$.

Question:

Is there a set $S\subseteq V$ such that $S$ is a size-at-most-$k$ vertex cover for $G$?

To prove Theorem 4.1(i), we give a polynomial-time many-one reduction from HCVC to