On Efficient Domination for Some Classes of H-Free Chordal Graphs

On Efficient Domination for Some Classes of -Free Chordal Graphs

Andreas Brandstädt
Institut für Informatik, Universität Rostock, D-18051 Rostock, Germany
andreas.brandstaedt@uni-rostock.de
   Raffaele Mosca
Dipartimento di Economia, Universitá degli Studi “G. D’Annunzio”, Pescara 65121, Italy
r.mosca@unich.it
Abstract

A vertex set in a finite undirected graph is an efficient dominating set (e.d.s. for short) of if every vertex of is dominated by exactly one vertex of . The Efficient Domination (ED) problem, which asks for the existence of an e.d.s. in , is known to be -complete even for very restricted graph classes such as for -free chordal graphs while it is solvable in polynomial time for -free chordal graphs (and even for -free graphs). A standard reduction from the -complete Exact Cover problem shows that ED is -complete for a very special subclass of chordal graphs generalizing split graphs. The reduction implies that ED is -complete e.g. for double-gem-free chordal graphs while it is solvable in linear time for gem-free chordal graphs (by various reasons such as bounded clique-width, distance-hereditary graphs, chordal square etc.), and ED is -complete for butterfly-free chordal graphs while it is solvable in linear time for -free graphs.

We show that (weighted) ED can be solved in polynomial time for -free chordal graphs when is net, extended gem, or .

Keywords: Weighted efficient domination; -free chordal graphs; -completeness; net-free chordal graphs; extended-gem-free chordal graphs; -free chordal graphs; polynomial time algorithm; clique-width.

1 Introduction

Let be a finite undirected graph. A vertex dominates itself and its neighbors. A vertex subset is an efficient dominating set (e.d.s. for short) of if every vertex of is dominated by exactly one vertex in ; for any e.d.s.  of , for every (where denotes the closed neighborhood of ). Note that not every graph has an e.d.s.; the Efficient Dominating Set (ED) problem asks for the existence of an e.d.s. in a given graph .

The Exact Cover Problem (X3C [SP2] in [17]) asks for a subset of a set family over a ground set, say , containing every vertex in exactly once. As shown by Karp [19], the problem is -complete even for set families containing only -element subsets of (see [17]).

Clearly, ED is the Exact Cover problem for the closed neighborhood hypergraph of . The notion of efficient domination was introduced by Biggs [3] under the name perfect code. The ED problem is motivated by various applications, including coding theory and resource allocation in parallel computer networks; see e.g. [1, 2, 3, 12, 20, 21, 22, 25, 26, 28, 29].

In [1, 2], it was shown that the ED problem is -complete. Moreover, ED is -complete for -free chordal unipolar graphs [14, 27, 29].

In this paper, we will also consider the following weighted version of the ED problem:

Weighted Efficient Domination (WED) Instance: A graph , vertex weights . Task: Find an e.d.s. of minimum finite total weight, or determine that contains no such e.d.s.

The relationship between WED and ED is analyzed in [6].

For a set of graphs, a graph is called -free if contains no induced subgraph isomorphic to a member of . In particular, we say that is -free if is -free. Let denote the disjoint union of graphs and , and for , let denote the disjoint union of copies of . For , let denote the chordless path with vertices, and let denote the complete graph with vertices (clearly, ). For , let denote the chordless cycle with vertices.

For indices , let denote the graph with vertices , , such that the subgraph induced by forms a , the subgraph induced by forms a , and the subgraph induced by forms a , and there are no other edges in . Thus, claw is , chair is , and is isomorphic to e.g. . is a linear forest if every component of is a chordless path, i.e., is claw-free and cycle-free.

is a co-chair if it is the complement graph of a chair. is a if has five vertices such that four of them induce a and the fifth is adjacent to exactly one of the -vertices. is a co- if is the complement graph of a . is a bull if has five vertices such that four of them induce a and the fifth is adjacent to exactly the two mid-points of the . is a net if has six vertices such that five of them induce a bull and the sixth is adjacent to exactly the vertex of the bull with degree 2. is a gem if has five vertices such that four of them induce a and the fifth is adjacent to all of the vertices. is a co-gem if is the complement graph of a gem.

For a vertex , denotes its (open) neighborhood, and denotes its closed neighborhood. A vertex sees the vertices in and misses all the others. The non-neighborhood of a vertex is . For , and .

We say that for a vertex set , a vertex has a join (resp., co-join) to if (resp., ). Join (resp., co-join) of to is denoted by (resp., ). Correspondingly, for vertex sets with , denotes for all and denotes for all . A vertex contacts if has a neighbor in . For vertex sets with , contacts if there is a vertex in contacting .

If but has neither a join nor a co-join to , then we say that distinguishes . A set of at least two vertices of a graph is called homogeneous if and every vertex outside is either adjacent to all vertices in , or to no vertex in . Obviously, is homogeneous in if and only if is homogeneous in the complement graph . A graph is prime if it contains no homogeneous set.

A graph is chordal if it is -free for any . is unipolar if can be partitioned into a clique and the disjoint union of cliques, i.e., there is a partition such that is a complete subgraph and is -free. is a split graph if and its complement graph are chordal. Equivalently, can be partitioned into a clique and an independent set. It is well known that is a split graph if and only if it is ()-free [15].

It is well known that ED is -complete for claw-free graphs (even for ()-free perfect graphs [24]) as well as for bipartite graphs (and thus for triangle-free graphs) [25] and for chordal graphs [14, 27, 29]. Thus, for the complexity of ED on -free graphs, the most interesting cases are when is a linear forest. Since (W)ED is -complete for -free graphs and polynomial for -free graphs [7], -free graphs were the only open case finally solved in [10, 11] by a direct polynomial time approach (and in [23] by an indirect one).

It is well known that for a graph class with bounded clique-width, ED can be solved in polynomial time [13]. Thus we only consider ED on -free chordal graphs for which the clique-width is unbounded. For example, the clique-width of -free chordal graphs is unbounded for claw-free chordal graphs while it is bounded if bull, gem, co-gem, co-chair. In [4], for all but two stubborn cases, the clique-width of -free chordal graphs is classified.

For graph , the square has the same vertex set , and two vertices , , are adjacent in if and only if . The WED problem on can be reduced to Maximum Weight Independent Set (MWIS) on (see [8, 6, 9, 26]).

While the complexity of ED for -free chordal graphs is -complete (as mentioned above), it was shown in [5] that WED is solvable in polynomial time for -free chordal graphs, since for -free chordal graphs with e.d.s., is chordal. It is well known [16] that MWIS is solvable in linear time for chordal graphs.

For -free chordal graphs, however, there are still many open cases. Motivated by the approach in [5], and the result of Milanič [26] showing that for (,net)-free graphs , its square is claw-free, we show in the next section that is chordal for -free chordal graphs with e.d.s. when is a net or an extended gem (see Figure 1 - extended gem generalizes ), and thus, WED is solvable in polynomial time for these two graph classes.

2 For Net-Free Chordal Graphs and Extended-Gem-Free Chordal Graphs with e.d.s., is Chordal

Figure 1: net and extended gem

Let be a chordal graph and its square.

Claim 2.1.

Let , , form a in with and , , (index arithmetic modulo ). Then

  • for each , ; let be a common neighbor of and in (an auxiliary vertex).

  • for each , , we have , and .

Proof. (i): If without loss of generality, then, since and , we have and ; let be a common neighbor of and be a common neighbor of . Clearly, since . Moreover, since and since . Now, since otherwise would induce a in but now in any case, the induced by leads to a chordless cycle in which is a contradiction.

(ii): Clearly, as above, we have for any , and a non-edge would lead to a chordless cycle in . ∎

Lemma 1.

If is a net-free chordal graph with e.d.s., then is chordal.

Proof. Let be a net-free chordal graph and assume that contains an e.d.s. .

Case 1. First suppose to the contrary that contains , say with vertices such that and , (index arithmetic modulo 4). By Claim 2.1, we have for each ; let denote a common neighbor of (an auxiliary vertex). By Claim 2.1, for . Since is chordal, either induce a diamond or in .

Assume first that induce a diamond in , say with .

Case 1.1 .

If then by the e.d.s. property, . Let with . Clearly, and , and by the e.d.s. property, , , , and since is chordal, but then induce a net in . Thus, and correspondingly, .

If then . Let with and with ; clearly, and . Since is chordal, (and in particular, and ). Clearly, by the e.d.s. property, , .

Now, if then induce a net in , and if then and now, induce a net in . Thus, and correspondingly, . This implies .

Case 1.2. and .

If without loss of generality, then . Let with and with . Clearly, since is chordal, we have and .

If then induce a net. Thus, . Let with . If and then induce a net. If then induce a net. Finally, if then induce a net. Thus, also Case 1.2 is excluded, and .

Case 1.3. .

Let be the -neighbor of ; . Clearly, since is chordal and since , are pairwise distinct.

If and then induce a net in , and correspondingly by symmetry, for , . Thus, we can assume that for each , sees at least one of .

If and then clearly, and and thus, by the above, we can assume that and but now, induce a net in .

Thus, assume that is adjacent to exactly one of , say (which implies ) and . By symmetry, this holds for as well, i.e., , , and . Then induce a net in .

In a very similar way, we can show that we can exclude a in when induce a in .

Case 2. Now suppose to the contrary that contains , , say with vertices such that and , , (index arithmetic modulo ). By Claim 2.1, we have for each ; let denote a common neighbor of . Again, by Claim 2.1, the auxiliary vertices are pairwise distinct and for each . Since and is chordal, there is an edge having a common neighbor , , say without loss of generality, induce a in . Then induce a net in (note that for , we do not need the existence of an e.d.s. in ).

Similarly, we get a net for any in , . Thus, Lemma 1 is shown. ∎

By [6], and since MWIS is solvable in linear time for chordal graphs [16], we obtain:

Corollary 1.

WED is solvable in time for net-free chordal graphs.

Lemma 1 generalizes the corresponding result for AT-free chordal graphs (i.e., interval graphs).

Lemma 2.

If is an extended-gem-free chordal graph with e.d.s., then is chordal.

Proof. Let be an extended-gem-free chordal graph and assume that contains an e.d.s. .

Case 1. First suppose to the contrary that contains , say with vertices such that and , (index arithmetic modulo 4). By Claim 2.1, we have for each ; let denote a common neighbor of (an auxiliary vertex). By Claim 2.1, for . Since is chordal, either induce a diamond or in .

Assume first that induce a diamond in , say with and .

Case 1.1 .

If then by the e.d.s. property, . Let with and with . Clearly, since is chordal, , and by the e.d.s. property, , and , but then induce an extended gem. Thus, and correspondingly, .

If then by the e.d.s. property, . Let with and . Clearly, since is chordal, , and in particular, and is nonadjacent to all neighbors of but then induce an extended gem. Thus, and correspondingly, .

Case 1.2. and .

Without loss of generality, assume that ; then by the e.d.s. property, . Let with and with .

First assume that . Then clearly, induce an extended gem. Thus, ; let with . Note that since and since otherwise, would induce a in but now again, induce an extended gem. Thus, also .

Case 1.3. .

For , let be the -neighbor of ; , . Clearly, since and since is chordal, are pairwise distinct.

By the e.d.s. property, or ; without loss of generality, assume that . Since do not induce a , we have , and correspondingly, but now, induce an extended gem.

Thus, is -free.

In a very similar way, we can show that we can exclude a in when induce a in .

Case 2. Now suppose to the contrary that contains , , say with vertices such that and , , (index arithmetic modulo ). By Claim 2.1, we have for each ; let denote a common neighbor of . Again, by Claim 2.1, the auxiliary vertices are pairwise distinct and for each .

Clearly, since is chordal, there is an edge .

Claim 2.2.

If then and .

Proof. Without loss of generality, let . If then clearly, and ; let with . Clearly, since but now, induce an extended gem. Thus, .

If then clearly, ; let with but now, induce an extended gem. Thus, and correspondingly, .

If then clearly, ; let with but now, induce an extended gem. Thus, and correspondingly, which shows Claim 2.2. ∎

Claim 2.3.

If then and .

Proof. Without loss of generality, let and suppose to the contrary that . Then by Claim 2.2, there are new vertices , , with and . By the e.d.s. property, or ; say without loss of generality, . Then by the chordality of , and but now, induce an extended gem. Thus, Claim 2.3 is shown. ∎

For a in , Claim 2.3 leads to a in induced by if . Thus, from now on, let .

Claim 2.4.

If then and .

Proof. Without loss of generality, let and suppose to the contrary that . Then by Claim 2.3, and as well as and , and since is chordal, and .

Since does not induce an extended gem, we have . For this contradicts the fact that . Thus, from now on, let .

Since does not induce an extended gem, we have . For , again, this contradicts the fact that . Thus, from now on, let .

Now, induce an extended gem. Thus, Claim 2.4 is shown. ∎

Now we can assume that ; without loss of generality, let . Then by Claims 2.3 and 2.4, we have , , and , . Since is chordal, we have .

Since does not induce an extended gem, we have .

Since does not induce an extended gem, we have (which, for contradicts the fact that ) but now, induce an extended gem.

Thus, Lemma 2 is shown. ∎

By [6], and since MWIS is solvable in linear time for chordal graphs [16], we obtain:

Corollary 2.

WED is solvable in time for extended-gem-free chordal graphs.

3 WED is -complete for a Special Class of Chordal Graphs

As mentioned in the Introduction, the reduction from X3C to Efficient Domination shows that ED is -complete. For making this manuscript self-contained, we describe the reduction here:

Let with and be a hypergraph with for all . Let be the following reduction graph:

such that , and are pairwise disjoint. The edge set of consists of all edges whenever . Moreover is a clique in , and every is only adjacent to .

Clearly, has an exact cover if and only if has an e.d.s. : For an exact cover of , every corresponds to vertex , and every corresponds to vertex . Conversely, if is an e.d.s. in we can assume without loss of generality that , and now, corresponds to an exact cover of .

Figure 2: , , , butterfly, extended butterfly, extended co-, extended chair, and double-gem

Clearly, is chordal and unipolar. The reduction shows that is not only -free but also -free for various other graphs such as , , butterfly, extended butterfly, extended co-, extended chair, and double-gem as shown in Figure 2; actually, it corresponds to a slight generalization of split graphs which was described by Zverovich in [30] as satgraphs.

Proposition 1.

ED is -complete for , , , butterfly, extended butterfly, extended co-, extended chair, double-gem-free chordal graphs

The reduction implies that ED is -complete e.g. for double-gem-free chordal graphs while it is solvable in linear time for gem-free chordal graphs (by various reasons such as bounded clique-width, distance-hereditary graphs, chordal square etc.), and ED is -complete for butterfly-free chordal graphs while it is solvable in linear time for -free graphs [9].

The clique-width of co--free chordal graphs, of -free chordal graphs, and of claw-free chordal graphs is unbounded (see [4]). Since co- and are subgraphs of extended gem, Lemma 2 implies the following result:

Lemma 3.

For -free chordal graphs and for co--free chordal graphs, WED is solvable in polynomial time.

Proof. As mentioned, Lemma 3 follows from Lemma 2. Here we give a direct proof of Lemma 3. First let be a connected -free chordal graph, and assume that contains a , say (otherwise is a tree since a -free chordal graph is a forest). Clearly, the non-neighborhood of is an independent vertex set, and according to the approach described in [7], it is clear that there are at most three vertices in an e.d.s.  dominating ; then the remaining non-dominated vertices in the non-neighborhood of belong to .

Another direct approach works as follows: If has an e.d.s.  such that for a , one of its vertices, say , is in then for distance levels of , is independent and thus, (recall that ), and it has to be checked whether is an e.d.s. of .

Now assume that no vertex of any in is in . Let again be a in , say with vertices . Then and is independent. Let be the -neighbors of , respectively, and let . Clearly, . Thus, for each possible triple of -vertices , one has to check whether is an e.d.s. of .

Now let be a prime co--free chordal graph (by [7], we can reduce the WED problem to prime graphs). As above, we can assume that contains a , say with vertices . If , say , then let be a vertex distinguishing an edge, say , in . Now, since , there is a -neighbor of but now induce a co- in which is a contradiction.

Thus assume that for every in , . This implies that any -vertex has at most one neighbor in any in . Thus assume that each vertex in has its personal -neighbor, say . If one of the -neighbors of , say, , has another neighbor then since otherwise, the induced by would contain a -vertex. Clearly, and since and and is chordal. Thus, would induce a co-