On Efficient Domination for Some Classes of Free Chordal Graphs
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 doublegemfree chordal graphs while it is solvable in linear time for gemfree chordal graphs (by various reasons such as bounded cliquewidth, distancehereditary graphs, chordal square etc.), and ED is complete for butterflyfree 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; netfree chordal graphs; extendedgemfree chordal graphs; free chordal graphs; polynomial time algorithm; cliquewidth.
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 clawfree and cyclefree.
is a cochair 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 midpoints 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 cogem 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 nonneighborhood of a vertex is . For , and .
We say that for a vertex set , a vertex has a join (resp., cojoin) to if (resp., ). Join (resp., cojoin) 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 cojoin 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 clawfree graphs (even for ()free perfect graphs [24]) as well as for bipartite graphs (and thus for trianglefree 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 cliquewidth, ED can be solved in polynomial time [13]. Thus we only consider ED on free chordal graphs for which the cliquewidth is unbounded. For example, the cliquewidth of free chordal graphs is unbounded for clawfree chordal graphs while it is bounded if bull, gem, cogem, cochair. In [4], for all but two stubborn cases, the cliquewidth 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 clawfree, 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 NetFree Chordal Graphs and ExtendedGemFree Chordal Graphs with e.d.s., is Chordal
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 nonedge would lead to a chordless cycle in . ∎
Lemma 1.
If is a netfree chordal graph with e.d.s., then is chordal.
Proof. Let be a netfree 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. ∎
Corollary 1.
WED is solvable in time for netfree chordal graphs.
Lemma 1 generalizes the corresponding result for ATfree chordal graphs (i.e., interval graphs).
Lemma 2.
If is an extendedgemfree chordal graph with e.d.s., then is chordal.
Proof. Let be an extendedgemfree 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. ∎
Corollary 2.
WED is solvable in time for extendedgemfree 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 selfcontained, 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 .
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 doublegem 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, doublegemfree chordal graphs
The reduction implies that ED is complete e.g. for doublegemfree chordal graphs while it is solvable in linear time for gemfree chordal graphs (by various reasons such as bounded cliquewidth, distancehereditary graphs, chordal square etc.), and ED is complete for butterflyfree chordal graphs while it is solvable in linear time for free graphs [9].
The cliquewidth of cofree chordal graphs, of free chordal graphs, and of clawfree 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 cofree 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 nonneighborhood 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 nondominated vertices in the nonneighborhood 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 cofree 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