Finding dominating induced matchings in -free graphs in polynomial time
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 . The notion of efficient domination was introduced by Biggs [1] under the name perfect code. The Efficient Domination (ED) problem asks for the existence of an e.d.s. in a given graph (note that not every graph has an e.d.s.)
A set of edges in a graph is an efficient edge dominating set (e.e.d.s. for short) of if and only if it is an e.d.s. in its line graph . The Efficient Edge Domination (EED) problem asks for the existence of an e.e.d.s. in a given graph . Thus, the EED problem for a graph corresponds to the ED problem for its line graph . Note that not every graph has an e.e.d.s. An efficient edge dominating set is also called dominating induced matching (d.i.m. for short)–recall that an edge subset is a dominating induced matching in if every edge in is intersected by exactly one edge of .
The EED problem is motivated by applications such as parallel resource allocation of parallel processing systems, encoding theory and network routing—see e.g. [9], [10]. The EED problem is called the Dominating Induced Matching (DIM) problem in various papers (see e.g. [2], [3], [8], [10], [11]); subsequently, we will use this terminology in our manuscript.
In [9], it was shown that the DIM problem is -complete; see e.g. [2], [8], [12], [13]. However, for various graph classes, DIM is solvable in polynomial time. For mentioning some examples, we need the following notions:
Let denote the path with vertices, say , and edges , . Such a path will also be denoted by . When speaking about a in a graph , we will always assume that the path is chordless, or, equivalently, that it is an induced subgraph of .
For indices , let denote the graph with vertices , , such that the subgraph induced by forms a , the subgraph induced by forms a , the subgraph induced by forms a , and there are no other edges in . Vertex is called the center of this . Thus, claw is , and is isomorphic to .
In [10], it is conjectured that for every fixed , DIM is solvable in polynomial time for -free graphs (actually, an even stronger conjecture is mentioned in [10]); this includes -free graphs for . The following results are known:
Theorem 1 DIM is solvable in polynomial time for -free graphs [8], -free graphs [11], -free graphs [10], -free graphs [4], -free graphs [7], -free graphs [3] (in this case even in linear time), -free graphs [5], and -free graphs [6].
Based on the two distinct approaches described in [5] and in [10], [11] (and combining them as in [4] and [7]), we show in this paper that DIM can be solved in polynomial time for -free graphs.
Section snippets
Basic notions
Let be a finite undirected graph without loops and multiple edges. Let denote its vertex set and its edge set; let and . For , let denote the open neighborhood of , and let denote the closed neighborhood of . The degree of in is . If , we also say that and see each other, and if , we say that and miss each other. A vertex set is independent in if for every pair of vertices , . A vertex set is a
Coloring
As in [4], we have:
Proposition 1 The following statements hold: For every edge , , with and (possibly ), we have . For every edge with and , holds, and thus, is an -forced -edge.
Proof : By (6), does not contain any -edge, and clearly, if then either or is black; without loss of generality, let be black but then and is white, i.e., . : By Observation 12 , and thus, by Proposition 1 , for the edge
Coloring
Recall that and . From now on, let . Subsequently, we apply the polynomial-time solution for -free graphs [10] (see Theorem 1 ). In particular let us try to connect the “coloring approach” of [10] with the above. Recall that for a d.i.m. of , is a partition of , all vertices of are black and all vertices of are white. Recall the following forcing rules (under the assumption that ):
- (i)
If a vertex is white then all of its neighbors
The facilitated case in which is empty
Lemma 3 If then one can detect a white vertex of in polynomial time.
Proof Assume . Then , hence with . If then since , i.e., but by Proposition 8, , which is a contradiction. Thus, . Then by the fixed -coloring of (if ) and by Proposition 7 (if ), the color of is fixed. Now, since induce a , we have: if is black and , then is white; if is black and , then is white; if is white, then is white.
The general case in which may be non-empty
By Proposition 10, let us distinguish between and . Then the goal is to detect a white vertex of or a peripheral triangle with a vertex of . Recall .
Proposition 12 If , then one can easily detect a white vertex of .
Proof In fact, if , then the color of (recall ) is known by Proposition 7. Then, as in the proof of Lemma 3, one can easily detect a white vertex of . □
From now on, let us assume that . In particular, let be a neighbor of , and let
A polynomial algorithm for DIM on -free graphs
The following procedure is part of the algorithm:
Procedure 7.1 Set . Determine the distance levels with respect to . Check whether is an independent set (see fact (2)) and is the disjoint union of edges and isolated vertices (see fact (3)). If not, then STOP with failure. For the set of edges in , apply the Edge Reduction for every edge in . Moreover, apply the Edge Reduction for each edge according to fact ((7)) and then for each edge according to Observation 12 . if DIM-with-
CRediT authorship contribution statement
Raffaele Mosca: Software, Validation, Writing - review & editing.
Acknowledgment
We gratefully thank the anonymous reviewers for their comments and corrections. The second author would like to witness that he just tries to pray a lot and is not able to do anything without that—ad laudem Domini.
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