Blocking dominating sets for H-free graphs via edge contractions

06/28/2019 ∙ by Esther Galby, et al. ∙ 0

In this paper, we consider the following problem: given a connected graph G, can we reduce the domination number of G by one by using only one edge contraction? We show that the problem is coNP-hard when restricted to subcubic claw-free graphs and P_7-free graphs.

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1 Introduction

A blocker problem asks whether given a graph , a graph parameter , a set of one or more graph operations and an integer , can be transformed into a graph by using at most operations from such that for some threshold . Such a designation follows from the fact that the set of vertices or edges involved can be viewed as ”blocking” the parameter . Identifying such sets may provide information on the structure of the input graph; for instance, if , and vertex deletion, the problem is equivalent to testing whether the input graph contains a vertex that is in every maximum independent set (see [17]). Blocker problems have received much attention in the recent literature (see for instance [1, 2, 3, 4, 5, 7, 8, 10, 11, 12, 14, 15, 16, 17, 18, 19]) and have been related to other well-known graph problems such as Hadwiger Number, Club Contraction and several graph transversal problems (see for instance [7, 16]). The graph parameters considered so far in the literature are the chromatic number, the independence number, the clique number, the matching number and the vertex cover number while the set is a singleton consisting of a vertex deletion, edge contraction, edge deletion or edge addition. In this paper, we focus on the domination number , let consists of an edge contraction and set the threshold to one.

Formally, let be a graph. The contraction of an edge removes vertices and from and replaces them by a new vertex that is made adjacent to precisely those vertices that were adjacent to or in (without introducing self-loops nor multiple edges). We say that a graph can be -contracted into a graph , if can be transformed into by a sequence of at most  edge contractions, for an integer . The problem we consider is then the following (note that contracting an edge cannot increase the domination number).

.99 -Edge Contraction()

    Instance: A connected graph .
Question: Can be -contracted into a graph such that ?

Reducing the domination number using edge contractions was first considered in [9]. The authors proved that for a connected graph such that , we have , where denotes the minimum number of edge contractions required to transform into a graph such that (note that if then is a No-instance for -Edge Contraction() independently of the value of ). Thus, if is a connected graph with , then is always a Yes-instance for -Edge Contraction() when . It was later shown in [8] that -Edge Contraction() is -hard for and so, restrictions on the input graph to some special graph classes were considered. In particular, the authors in [8] proved that for , the problem is polynomial-time solvable for -free graphs while for , it remains -hard when restricted to -free graphs and -free graphs, for any .

In this paper, we continue the systematic study of the computational complexity of 1-Edge Contraction() initiated in [8]. Ultimately, the aim is to obtain a complete classification for 1-Edge Contraction() restricted to -free graphs, for any (not necessarily connected) graph , as it has been done for other blocker problems (see for instance [17, 18, 19]). As a step towards this end, we prove the following two theorems.

Theorem 1.1.

1-Edge Contraction() is -hard when restricted to subcubic claw-free graphs.

Theorem 1.2.

1-Edge Contraction() is -hard when restricted to -free graphs.

Since 1-Edge Contraction() is -hard when restricted to -free graphs, for any , it follows that 1-Edge Contraction() is -hard for -free graphs when contains a cycle. If is a forest with a vertex of degree at least three, we conclude by Theorem 1.1 that 1-Edge Contraction() is -hard for -free graphs; and if is a linear forest containing a path of length at least 7, then 1-Edge Contraction() is -hard for -free graphs by Theorem 1.2. There remains to determine the complexity status of the problem restricted to -free graphs when is a disjoint union of paths of length at most 6, which we leave as an open problem.

2 Preliminaries

Throughout the paper, we only consider finite, undirected, connected graphs that have no self-loops or multiple edges. We refer the reader to [6] for any terminology and notation not defined here.

For , the path and cycle on vertices are denoted by and respectively. The claw is the complete bipartite graph with one partition of size one and the other of size three.

Let be a graph and let . We denote by , or simply if it is clear from the context, the set of vertices that are adjacent to i.e., the neighbors of , and let . The degree of a vertex , denoted by or simply if it is clear from the context, is the size of its neighborhood i.e., . The maximum degree in is denoted by and is subcubic if .

For any graph , is said to be -free if contains no induced subgraph isomorphic to . For a subset , we let denote the subgraph of induced by , which has vertex set  and edge set .

A subset is called an independent set or is said to be independent, if no two vertices in are adjacent. A subset is called a dominating set, if every vertex in is adjacent to at least one vertex in ; the domination number is the number of vertices in a minimum dominating set. For any and , is said to dominate (in particular, dominates itself). We say that contains an edge (or more) if the graph contains an edge (or more). A dominating set of is efficient if for every vertex , that is, is dominated by exactly one vertex.

In the following, we consider those graphs for which one contraction suffices to decrease their domination number by one. A characterization of this class is given in [9].

Theorem 2.1 ([9]).

For a connected graph , if and only if there exists a minimum dominating set in that is not independent.

In order to prove Theorems 1.1 and 1.2, we introduce to two following problems.

.99 All Efficient MD

[2pt]     Instance: A connected graph . Question: Is every minimum dominating set of efficient?

.99 All Independent MD

[2pt]     Instance: A connected graph . Question: Is every minimum dominating set of independent?

The following is then a straightforward consequence of Theorem 2.1.

Fact 2.2.

Given a graph , is a Yes-instance for 1-Edge Contraction() if and only if is a No-instance for All Independent MD.

3 The proof of Theorem 1.1

In this section, we show that 1-Edge Contraction() is -hard when restricted to subcubic claw-free graphs. To this end, we first prove the following.

Lemma 3.1.

All Efficient MD is -hard when restricted to subcubic graphs.

Proof.

We reduce from Positive Exactly 3-Bounded 1-In-3 3-Sat, where each variable appears in exactly three clauses and only positively, each clause contains three positive literals, and we want a truth assignment such that each clause contains exactly one true literal. This problem is shown to be -complete in [13]. Given an instance of this problem, with variable set and clause set , we construct an equivalent instance of All Efficient MD as follows. For any variable , we introduce a copy of , which we denote by , with three distinguished true vertices , and , and three distinguished false vertices , and (see Fig. 0(a)). For any clause containing variables , and , we introduce the gadget depicted in Fig. 0(b) which has one distinguished clause vertex and three distinguished variable vertices , and (note that is not connected). For every , we then add an edge between and and between and for some so that (resp. ) is adjacent to exactly one variable vertex (resp. clause vertex). We denote by the resulting graph. Note that .

(a) The variable gadget .

(b) The clause gadget .
Figure 1: Construction of the graph (the rectangle indicates that the corresponding set of vertices induces a clique).
Observation 3.1.1.

For any dominating set of , for any and for any . In particular, .

Indeed, for any , since , and must be dominated and their neighborhoods are pairwise disjoint and contained in , it follows that . For any , since the vertices of must be dominated and their neighborhoods are contained in , .

Observation 3.1.2.

For any , if is a minimum dominating set of then either , or .

Claim 3.1.3.

is satisfiable if and only if .

Proof.

Assume that is satisfiable and consider a truth assignment satisfying . We construct a dominating set of as follows. For any variable , if is true, add , and to ; otherwise, add , and to . For any clause containing variables , and , exactly one variable is true, say without loss of generality; we then add to . Clearly, is dominating and we conclude by Observation 3.1.1 that .

Conversely, assume that and consider a minimum dominating set of . Then by Observation 3.1.1, for any and for any . Now, for a clause containing variables , and , if then and so, at least two vertices from are not dominated; thus, . It follows that for any , is a minimum dominating set of which by Observation 3.1.2 implies either or ; and we conclude similarly that either or . Now given a clause containing variables , and , since , at least one true vertex adjacent to the clause vertex must belong to , say for some without loss of generality. It then follows that and which implies that (either or a vertex from would otherwise not be dominated). But then, since for , must be dominated, it follows that . We thus construct a truth assignment satisfying as follows: for any variable , if , set to true, otherwise set to false. ∎

Claim 3.1.4.

if and only if every minimum dominating set of is efficient.

Proof.

Assume that and consider a minimum dominating set of . Then by Observation 3.1.1, for any and for any . As shown previously, it follows that for any clause containing variables , and , ; and for any , either or (we conclude similarly with and ). Thus, for any , every vertex in is dominated by exactly one vertex. Now given a clause containing variables , and , since the clause vertex does not belong to , there exists at least one true vertex adjacent to which belongs to . Suppose to the contrary that has strictly more than one neighbor in , say and without loss of generality. Then, for which implies that as for . It follows that the variable vertices and must be dominated by some vertices in ; but and and so, either or is not dominated. Thus, has exactly one neighbor in , say without loss of generality. Then, necessarily for otherwise either or some vertex in would not be dominated. But then, it is clear that every vertex in is dominated by exactly one vertex; thus, is efficient.

Conversely, assume that every minimum dominating set of is efficient and consider a minimum dominating set of . If for some , , then clearly at least one vertex in is dominated by two vertices in . Thus, for any and we conclude by Observation 3.1.1 that in fact, equality holds. The next observation immediately follows from the fact that is efficient.

Observation 3.1.5.

For any , if then either , or .

Now, consider a clause containing variables , and and suppose without loss of generality that is adjacent to (note that then the variable vertex is adjacent to ). If the clause vertex belongs to then, since is efficient, and ( would otherwise be dominated by at least two vertices) which contradicts Observation 3.1.5. Thus, no clause vertex belongs to . Similarly, suppose that there exists such that , say without loss of generality. Then, since is efficient, and ( would otherwise be dominated by at least two vertices) which again contradicts Observation 3.1.5. Thus, no variable vertex belongs to . Finally, since is efficient, and so, by Observation 3.1.1. ∎

Now by combining Claims 3.1.3 and 3.1.4, we obtain that is satisfiable if and only if every minimum dominating set of is efficient, that is, is a Yes-instance for All Efficient MD. ∎

Theorem 3.2.

All Independent MD is -hard when restricted to subcubic claw-free graphs.

Proof.

We use a reduction from Positive Exactly 3-Bounded 1-In-3 3-Sat, where each variable appears in exactly three clauses and only positively, each clause contains three positive literals, and we want a truth assignment such that each clause contains exactly one true literal. This problem is shown to be -complete in [13]. Given an instance of this problem, with variable set and clause set , we construct an equivalent instance of All Independent MD as follows. Consider the graph constructed in the proof of Lemma 3.1 and let for (note that no vertex in has degree one). Then, for any , we replace the vertex by the gadget depicted in Fig. 1(a); and for any , we replace the vertex by the gadget depicted in Fig. 1(b). We denote by the resulting graph. Note that is claw-free and (also note that no vertex in has degree one). It is shown in the proof of Lemma 3.1 that is satisfiable if and only if is a Yes-instance for All Efficient MD; we here show that is a Yes-instance for All Efficient MD if and only if is a Yes-instance for All Independent MD. To this end, we first prove the following.

(a) .

(b) .
Figure 2: The gadget .
Claim 3.2.1.

.

Proof.

Let be a minimum dominating set of . We construct a dominating set of as follows. For any , if , add , , , , , and to ; otherwise, add , and to . For any , let be a neighbor of , say without loss of generality. Then, if , add , , , and to ; otherwise, add and to . Clearly, is dominating and .

Observation 3.2.2.

For any dominating set of , the following holds.

  • For any , . Moreover, if equality holds then and there exists such that .

  • For any , . Moreover, if equality holds then and there exists such that .

(i) Clearly, and as and must be dominated. Thus, . Now, suppose that say without loss of generality. Then as must be dominated which implies that (recall that ). Similarly, if both and belong to , then as ( would otherwise not be dominated).

(ii) Clearly, for any , as must be dominated. Now, if there exists such that , say without loss of generality, then (one of and would otherwise not be dominated). But then, as must be dominated, and as must be dominated; and so, (recall that ). Otherwise, for any , which implies that .

Now suppose that , say without loss of generality. If there exists such that , say without loss of generality, then (one of and would otherwise not be dominated). But then, as should be dominated, and as must be dominated. Since , it then follows that . Otherwise, for any and so, (recall that for any ).

Observation 3.2.3.

If is a minimum dominating set of , then for any and for any .

Indeed, if then is a dominating set of ; and if , then is a dominating set of .

Now, consider a minimum dominating set of and let and . We claim that is a dominating set of . Indeed, consider a vertex . We distinguish two cases depending on whether of .

Case 1. . Then by construction, which by Observation 3.2.2(i) implies that there exists such that , say without loss of generality. Since must be dominated, must then have a neighbor belonging to , for some vertex adjacent to in . But then, it follows from Observation 3.2.2 that if , and if (indeed, ); thus, .

Case 2. . Then by construction, which by Observation 3.2.2(ii) implies that there exists such that , say without loss of generality. Since must be dominated, must then have a neighbor belonging to , for some vertex adjacent to in . But then, it follows from Observation 3.2.2 that if , and if (indeed, ); thus, .

Hence, is a dominating set of . Moreover, it follows from Observations 3.2.2 and 3.2.3 that . Thus, and so, . Finally note that this implies that the constructed dominated set is in fact minimum. ∎

We next show that is a Yes-instance for All Efficient MD if and only if is a Yes-instance for All Independent MD. Since is satisfiable if and only if is a Yes-instance for All Efficient MD, as shown in the proof of Lemma 3.1, this would conclude the proof.

Assume first that is a Yes-instance for All Efficient MD and suppose to the contrary that is a No-instance for All Independent MD that is, has a minimum dominating set which is not independent. Denote by the minimum dominating set of constructed from according to the proof of Claim 3.2.1. Let us show that is not efficient. Consider two adjacent vertices . If and belong to gadgets and respectively, for two adjacent vertices and in , that is, is of the form and is of the form , then by Observation 3.2.2 and so, is not efficient. Thus, it must be that and both belong the same gadget , for some . We distinguish cases depending on whether or .

Case 1. . Suppose that . Then by Observation 3.2.2(i), and there exists such that , say without loss of generality. Then, necessarily ( would otherwise not be dominated) and so, as contains an edge and by assumption; but then, is not dominated. Thus, and we conclude by Observation 3.2.3 that in fact, equality holds. Note that consequently, . We claim that then, . Indeed, if both and belong to , then (since , would otherwise not be dominating) which contradicts that fact that contains an edge. Thus, and we may assume without loss of generality that . Let be the other neighbor of in , where is a neigbhor of in .

Suppose first that . Then, for otherwise would belong to and so, would contain the edge . It then follows from Observation 3.2.2(i) that there exists such that , where is the neighbor of in . We claim that ; indeed, if , since , would not be dominated. But then, must have a neighbor , for some vertex adjacent to in , which belongs to ; it then follows from Observation 3.2.2 and the construction of that and so, has two neighbors in , namely and , a contradiction.

Second, suppose that . Then, for otherwise would belong to and so, would contain the edge . It then follows from Observation 3.2.2(ii) that there exists such that , where and are the two neighbors of in . We claim that ; indeed, if , since , would not be dominated. But then, must have a neighbor , for some vertex adjacent to in , which belongs to ; it then follows from Observation 3.2.2 and the construction of that and so, has two neighbors in , namely and , a contradiction.

Case 2. . Suppose that . Then, by Observation 3.2.2(ii), and there exists such that , say without loss of generality. Then, (one of and would otherwise not be dominated), ( would otherwise not be dominated), ( would otherwise not be dominated) and ( would otherwise not be dominated); in particular, as by assumption. Since contains an edge, it follows that either or