Reducing the domination number of P_3+kP_2-free graphs via one edge contraction

10/27/2020 ∙ by Esther Galby, et al. ∙ 0

In this note, we consider the following problem: given a connected graph G, can we reduce the domination number of G by using only one edge contraction? We show that the problem is polynomial-time solvable on P_3+kP_2-free graphs for any k ≥ 0 which combined with results of [1,2] leads to a complexity dichotomy of the problem on H-free graphs.

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1 Introduction

Given a graph , a dominating set of is a subset of vertices such that every vertex in has a neighbor in , and is minimum if it has minimum cardinality amongst all dominating sets of . The domination number of is the cardinality of a minimum dominating set of . The contraction of an edge removes vertices and from and replaces them with a new vertex which is made adjacent to precisely those vertices that were adjacent to or in (without introducing self-loops nor multiple edges). In this note, we consider the following problem.

.99 1-Edge Contraction()

[2pt]     Instance: A connected graph . Question: Does there exist an edge such that contracting reduces the domination number by at least one?

It was shown in [beith] that the problem is -hard in general graphs. As a consequence, the authors considered restrictions of the input to special graph classes and proved in particular the following. [[beith], Corollary 1.4] is - or -hard on -free graphs if is not a linear forest111A linear forest is a forest of maximum degree 2, or equivalently, a disjoint union of paths. or contains an induced , or , and solvable in polynomial time if is an induced subgraph of for some . To obtain a complexity dichotomy of on -free graphs, there only remains to settle the complexity of the problem when is an induced subgraph of for some and . In this paper we solve this remaining case and prove the following. is polynomial-time solvable on -free graphs for any . Combined with results of [beith, dagstuhl], we obtain the following dichotomy. is polynomial-time solvable on -free graphs if and only if is an induced subgraph of for some or an induced subgraph of for some , unless .

2 Preliminaries

Throughout this paper the considered graphs are finite, simple and connected, unless stated otherwise. For , the path on vertices is denoted by . If is a graph and , then we denote by the graph consisting of disjoint copies of . Given a graph , we denote by its vertex set and by its edge set. The (open) neighbourhood of a vertex is the set . The closed neighbourhood of a vertex is the set . If and then we say that is complete to if is adjacent to every vertex in . For we write for the graph induced by , that is, the graph with vertex set and edge set . A set is called a clique (respectively a stable set) if every two vertices in are adjacent (respectively non-adjacent). For two vertices the distance from to is the number of edges in any shortest path between and . Given a dominating set of and a vertex , every vertex such that is called a private neighbour of . The following theorem characterises the -instances for . [[HuangXu], Lemma 3.5] A graph is a -instance for if and only if there exists a minimum dominating set of which is not a stable set. We will also use the following theorem which presents some cases where is polynomial-time solvable. [[dagstuhl], Proposition 12] can be solved in polynomial time for a graph class if for every graph we have , where is some fixed constant. If is a graph and can be solved in polynomial time for -free graphs, then  can be solved in polynomial time for -free graphs.

3 Proof of creftypecap 1

First observe that if a graph does not contain an induced then is a clique and thus a -instance for . Assume henceforth that and let be a -free containing an induced . Let be such that is isomorphic to , let be the set of vertices at distance one from and let the set of vertices at distance two from . Note that since is -free, the sets and partition and is a stable set. Denote by the set of vertices whose neighbourhoods are cliques. We call a vertex a regular vertex if there exist vertices such that are pairwise at distance at least four from one another. We denote by the set of regular vertices. Let be a regular vertex. If a vertex is adjacent to a vertex in then there exists a regular vertex such that is complete to .

Proof.

Let be regular vertices which are pairwise at distance at least four from one another. Suppose for a contradiction that is adjacent to and for every , there exists a vertex such that is not adjacent to . Then induces a , a contradiction. ∎

Let be a minimum dominating set of and let be a regular vertex. Then .

Proof.

Let be regular vertices which are pairwise at distance at least four from one another. As for any must be dominated,

Suppose for a contradiction that . For every , let be a vertex adjacent to . If a vertex is adjacent to for some then either in which case is adjacent or identical to , or and by creftypecap 3 there exists such that is complete to ; in particular, is then adjacent to . It now follows that is a dominating set of of cardinality less than , a contradiction to the minimality of . ∎

By using similar arguments as in the proof of creftypecap 3, we can prove the following. Let be regular vertices which are pairwise at distance at least four from one another. For every , let be a neighbour of . If is a minimum dominating set of then is a minimum dominating set of as well. No two regular vertices have a common neighbour. Furthermore, if there are two regular vertices at distance three from each other then is a -instance for .

Proof.

Suppose that there exist two regular vertices at distance at most three from one another. Let (respectively ) be regular vertices such that (respectively ) are pairwise at distance at least four from one another. For every , let (respectively ) be a neighbour of (respectively ) such that whenever . Suppose for a contradiction that and let be a common neighbour of and . Observe that cannot be adjacent to for any , since the neighbourhood of is a clique and is not adjacent to (recall that are pairwise at distance at least four from one another). It follows that the vertices induce a , a contradiction. Suppose now that and let and be two adjacent vertices. Consider a minimum dominating set of . Then by creftypecap 3 is a minimum dominating set of containing an edge; creftypecap 2 then implies the claim. ∎

Assume that is a -instance for and let be a minimum dominating set of . If there exists a vertex which has more than one private neighbour in then .

Proof.

Observe first that is a stable set by creftypecap 2. Assume that there exists a vertex which has at least two private neighbours in , say and . Suppose for a contradiction that there are at least further vertices in besides , say . We claim that for every there exists a vertex such that . Indeed, if for some there is no such vertex in then is a minimum dominating set of containing an edge, a contradiction to creftypecap 2. Now assume, without loss of generality, that is adjacent to for every . Then the vertices induce a , a contradiction. ∎

Assume that is a -instance for and let be a minimum dominating set. If there exists a vertex such that then is adjacent to all the vertices in except for at most .

Proof.

Observe first that is a stable set by creftypecap 2. Assume that has at least two neighbours in , say and . Suppose for a contradiction that there are at least vertices in which are not adjacent to , say . As shown in the proof of creftypecap 3, there has to be for every a vertex such that . Assume, without loss of generality, that is adjacent to for every . Then the vertices induce a , a contradiction. ∎

Assume that is a -instance for and let be a minimum dominating set of . If there are at least vertices in which do not have a private neighbour in then .

Proof.

Assume that . Suppose for a contradiction that there are at least vertices in , say , which have no private neighbours in . Then for every , any vertex has to be adjacent to at least two vertices in (note indeed that by creftypecap 2 does not belong to ) and thus by creftypecap 3, has to be adjacent to at least vertices in . But then is a minimum dominating set of containing an edge, a contradiction to creftypecap 2. ∎

Assume that is a -instance for and let be a minimum dominating set of . If there exists a vertex which has a private neighbour and a private neighbour such that is not adjacent to then .

Proof.

If a vertex in has two private neighbours in then we conclude by creftypecap 3. Thus, we can assume that no vertex in has more than one private neighbour in . Assume that has exactly one private neighbour and assume further that has a private neighbour such that and are not adjacent. Suppose for a contradiction that . Then by creftypecap 3 there are at most vertices in which do not have a private neighbour in . Hence, besides , there are at least further vertices in which do have private neighbours in . Let be such vertices with private neighbours , respectively. By the pigeonhole principle, there are either indices such that is non-adjacent to or indices such that is adjacent to . In the first case, assume, without loss of generality, that are non-adjacent to . Then the vertices induce a (recall that by creftypecap 2 is a stable set), a contradiction. In the second case, assume, without loss of generality, that is complete to . Then by creftypecap 3, every vertex in which is adjacent to a vertex in is adjacent to a vertex in as well. Thus, is a minimum dominating set of containing an edge, a contradiction to creftypecap 2. ∎

If is a -instance for then there exists a minimum dominating set of such that .

Proof.

Let be a minimum dominating set of such that is minimal amongst all minimum dominating sets of . If , then by creftypecap 3 every vertex in has at most one private neighbour in and creftypecap 3 ensures that there is at least one vertex which does have a private neighbour . But now either is a minimum dominating set of , contradicting the fact that is minimal amongst all minimum dominating sets of , or has a private neighbour which is not adjacent to , a contradiction to creftypecap 3. Hence and since the claim follows. ∎

Assume that is a -instance for and let be a minimum dominating set of . If is a subset of vertices which are pairwise at distance at least three from one another and every vertex in has two non-adjacent neighbours then .

Proof.

Assume first that is a set of vertices which are pairwise at distance at least three from one another and for every there are two non-adjacent vertices . If for every the vertices and were at distance at least four then the vertices would induce a , a contradiction. Hence there are two indices such that and are at distance exactly three from one another. Now suppose for a contradiction that there is a set of at least vertices which are pairwise at distance at least three from one another and such that for every vertex there are two vertices in which are not adjacent. By the above, there must exist two vertices in at distance exactly three. Let be a maximum subset of such that contains exactly one edge and no two vertices in share a common neighbour in . Observe that and that induces a . This implies in particular that . We construct a sequence of sets of vertices according to the following procedure.

  • Initialize . Set and .

  • Increase by one.

  • Let be a maximum set of vertices such that contains exactly one edge and no two vertices in share a common neighbour in . Set and .

  • If , stop the procedure. Otherwise, return to step 2.

Consider the value of at the end of the procedure (note that ). Observe that since for any , and , it follows that for any , . Let us show that . Since for any , , we have that . Thus if then for any , which implies by the above that for any . We now claim that cannot be larger than . Indeed, if then for any , with as shown previously; but for any which implies that , a contradiction. Thus and so . Now observe that for any vertex , every neighbour has to be adjacent to as otherwise the procedure would have output instead of . Furthermore, for any vertex every neighbour has to be adjacent to a vertex in as otherwise the procedure would have output instead of (recall that ). It follows that is a minimum dominating set of containing an edge, a contradiction to creftypecap 2. ∎

Assume that is a -instance for and let be a minimum dominating set of . Then the number of vertices in which are at distance two from another vertex in is at most .

Proof.

If every two vertices in are at distance at least three from one another then we are done. Thus assume that there are two vertices in which are at distance two from one another. Let and let be a set of minimum size in . Note that since there are two vertices in which have a common neighbour, is non-empty. If then is a dominating set of of cardinality at most which contains an edge, a contradiction to creftypecap 2. Hence . We now claim that every vertex in is adjacent to two vertices in which are not adjacent to any other vertex in . Indeed, if a vertex has no neighbour in which is not adjacent to any other vertex in then we could remove from without changing the cardinality of , thereby contradicting the fact that . If a vertex has only one neighbour in which is not adjacent to any other vertex in then removing from would only remove from , thus leaving the value of unchanged while decreasing the cardinality of , a contradiction to minimality of . This implies in particular that which combined with the inequality above leads to and . Now denote by the set of all vertices in which are not adjacent to a vertex in . Observe that no wo vertices can have a common neighbour , as otherwise would be such that and and thus , thereby contradicting the fact that . Thus every two vertices in are at distance at least three from one another and so by creftypecap 3, at most vertices in do not have cliques as neighbourhoods. Denote by the set of vertices whose neighbourhoods are cliques. Observe that no vertex in can be at distance two to another vertex in as otherwise we could remove from and replace it with a common neighbour of and , yielding a minimum dominating set of containing an edge, a contradiction to creftypecap 2. Thus, every vertex in which has a common neighbour with another vertex in must be contained in or in , which together have cardinality at most . ∎

If is a -instance for then there exists a minimum dominating set of such that .

Proof.

It follows from creftypecap 3 that there is a minimum dominating set such that . Let be the set of the vertices in which are at distance at least three to every other vertex in . Let be the set of the vertices in whose neighbourhoods are cliques. Suppose for a contradiction that there are two vertices which are at distance three. Let and be two adjacent vertices. Then is a minimum dominating set containing an edge, a contradiction to creftypecap 2. Thus, the vertices in are pairwise at distance at least four from one another. It follows that either or . Since by creftypecap 3 and by creftypecap 3 , the claim follows. ∎

We now present an algorithm which determines in polynomial time whether is a -instance of or not. In the following, we let .

  • Determine , , and .

    • If , check if there exists a dominating set of size at most .

    • Else go to 2.

  • Check whether there exist two regular vertices in which are at distance at/. most three from one another. If so, output .

  • Let be the set of vertices at distance exactly one from and let . If , output .

  • Determine . If , output .

  • Let be the family of all sets in of minimum size.

    • If there exists a set containing an edge, output .

    • If there exists a set such that , output .

  • Output .

Finally, let us show that this algorithm outputs the correct answer. In case then by creftypecap 3 is a -instance for if there exists no dominating set of size at most (see step 1.1.1). If such a set exists, then we conclude using creftypecap 2 (see step 1.1.2). If in step 2, two regular vertices at distance at most three from one another are found then by creftypecap 3, is a -instance for . Otherwise, any two regular vertices are at distance at least four from one another and by creftypecap 3, there exists a minimum dominating set of such that for any regular vertex , where . In the following, we denote by . Note that by creftypecap 3, for any , . Now if , then we conclude by creftypecap 3 and the fact that any two regular vertices are at distance at least four from one another, that any minimum total dominating set of is a stable set, that is, is a -instance for (see step 3). Otherwise and if is a -instance for , then by creftypecap 3 there must exist a set of cardinality at most such that for any , . Thus, if then is a -instance for (see step 4). Otherwise , and for any , is a minimum dominating set of . It then follows from creftypecap 2 that if there exists such that contains an edge then is a -instance for (see step 5(i)); otherwise, any is a stable set and if there exists a set such that then contains an edge and so, is a -instance by creftypecap 2 (see step 5(ii)). Otherwise, for any , is a stable set and which implies that is a stable set and thus, is a -instance for . As every step can clearly be done in polynomial time, this concludes the proof of creftypecap 1.