# Finding Dominating Induced Matchings in S_1,1,5-Free Graphs in Polynomial Time

Let G=(V,E) be a finite undirected graph. An edge set E' ⊆ E is a dominating induced matching ( d.i.m.) in G if every edge in E is intersected by exactly one edge of E'. The Dominating Induced Matching (DIM) problem asks for the existence of a d.i.m. in G; this problem is also known as the Efficient Edge Domination problem; it is the Efficient Domination problem for line graphs. The DIM problem is -complete even for very restricted graph classes such as planar bipartite graphs with maximum degree 3 but is solvable in linear time for P_7-free graphs, and in polynomial time for S_1,2,4-free graphs as well as for S_2,2,2-free graphs and for S_2,2,3-free graphs. In this paper, combining two distinct approaches, we solve it in polynomial time for S_1,1,5-free graphs.

## Authors

• 7 publications
• 8 publications
• ### Finding Dominating Induced Matchings in P_9-Free Graphs in Polynomial Time

Let G=(V,E) be a finite undirected graph. An edge set E' ⊆ E is a domin...
08/02/2019 ∙ by Andreas Brandstädt, et al. ∙ 0

• ### Some Results on Dominating Induced Matchings

Let G be a graph, a dominating induced matching (DIM) of G is an induced...
12/01/2019 ∙ by Saieed Akbari, et al. ∙ 0

• ### Efficient algorithms for maximum induced matching problem in permutation and trapezoid graphs

We first design an 𝒪(n^2) solution for finding a maximum induced matchin...
07/18/2021 ∙ by Viet Dung Nguyen, et al. ∙ 0

• ### Blocking dominating sets for H-free graphs via edge contractions

In this paper, we consider the following problem: given a connected grap...
06/28/2019 ∙ by Esther Galby, et al. ∙ 0

• ### Algorithms and Complexity on Indexing Founder Graphs

We study the problem of matching a string in a labeled graph. Previous r...
02/25/2021 ∙ by Massimo Equi, et al. ∙ 0

• ### Upper paired domination versus upper domination

A paired dominating set P is a dominating set with the additional proper...

• ### Budgeted Dominating Sets in Uncertain Graphs

We study the Budgeted Dominating Set (BDS) problem on uncertain graphs, ...
07/07/2021 ∙ by Keerti Choudhary, et al. ∙ 0

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## 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 . 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), and the EED problem is called the Dominating Induced Matching (DIM) problem in various papers (see e.g. [2, 3, 4, 5, 6, 7, 9, 10]); subsequently, we will use this notation instead of EED.

In [8], it was shown that the DIM problem is -complete; see also [2, 7, 11, 12]. However, for various graph classes, DIM is solvable in polynomial time. For mentioning some examples, we need the following notions:

Let denote the chordless path with vertices, say , and edges , ; we also denote it as .

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 ; is called the center of . Thus, claw is , and is isomorphic to .

For a set of graphs, a graph is called -free if no induced subgraph of is contained in . If , say , then instead of -free, is called -free.

The following results are known:

###### Theorem 1.

DIM is solvable in polynomial time for

• -free graphs [7],

• -free graphs [10],

• -free graphs [9],

• -free graphs [5],

• -free graphs [6],

• -free graphs [3] (in this case even in linear time),

• -free graphs [4].

In [9], it is conjectured that for every fixed , DIM is solvable in polynomial time for -free graphs (actually, an even stronger conjecture is mentioned in [9]); this includes -free graphs for .

Based on the two distinct approaches described in [4] and in [9, 10], we show in this paper that DIM can be solved in polynomial time for -free graphs (generalizing the corresponding result for -free graphs).

## 2 Definitions and Basic Properties

### 2.1 Basic notions

Let be a finite undirected graph without loops and multiple edges. Let or denote its vertex set and or its edge set; let and . For , let denote the open neighborhood of , and let denote the closed neighborhood of . 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 clique in if for every pair of vertices , , . For let and .

For , let denote the subgraph of induced by vertex set . Clearly is an edge in exactly when and ; thus, can simply be denoted by (if understandable).

For and , , we say that ( and miss each other) if there is no edge between and , and and see each other if there is at least one edge between and . If a vertex has a neighbor then contacts . If every vertex in sees every vertex in , we denote it by . For , we simply denote by , and correspondingly by . If for , , we say that is isolated in . For graphs , with disjoint vertex sets, denotes the disjoint union of , , and for , denotes the disjoint union of copies of . For example, is the disjoint union of two edges.

As already mentioned, a chordless path , , has vertices, say , and edges , ; the length of is . We also denote it as .

A chordless cycle , , has vertices, say , and edges , , and ; the length of is .

Let , , denote the clique with vertices. Let or diamond be the graph with four vertices, say , such that forms a and ; its mid-edge is the edge .

A butterfly has five vertices, say, , such that induce a with edges and (the peripheral edges of the butterfly), and .

We often consider an edge to be a set of two vertices; then it makes sense to say, for example, and , for an edge . For two vertices , let denote the distance between and in , i.e., the length of a shortest path between and in . The distance between a vertex and an edge is the length of a shortest path between and , i.e., . The distance between two edges is the length of a shortest path between and , i.e., . In particular, this means that if and only if .

An edge subset is an induced matching if the pairwise distance between its members is at least 2, that is, is isomorphic to for . Obviously, if is a d.i.m. then is an induced matching.

Clearly, has a d.i.m. if and only if every connected component of has a d.i.m.; from now on, connected components are mentioned as components.

Note that if has a d.i.m. , and denotes the vertex set of then is an independent set, say , i.e.,

 V has the partition V=V(M)∪I. (1)

From now on, all vertices in are colored white and all vertices in are colored black. According to [9], we also use the following notions: A partial black-white coloring of is feasible if the set of white vertices is an independent set in and every black vertex has at most one black neighbor. A complete black-white coloring of is feasible if the set of white vertices is an independent set in and every black vertex has exactly one black neighbor. Clearly, is a d.i.m. of if and only if the black vertices and the white vertices form a complete feasible coloring of .

### 2.2 Reduction steps, forbidden subgraphs, forced edges, and excluded edges

Various papers on this topic introduced and applied some forcing rules for reducing the graph to a subgraph such that has a d.i.m. if and only if has a d.i.m., based on the condition that for a d.i.m. , has the partition such that all vertices in are black and all vertices in are white (recall (1)).

A vertex is forced to be black if for every d.i.m.  of , . Analogously, a vertex is forced to be white if for every d.i.m.  of , .

Clearly, if and if are forced to be black, then is contained in every (possible) d.i.m. of .

An edge is a forced edge of if for every d.i.m.  of , . Analogously, an edge is an excluded edge of if for every d.i.m.  of , .

For the correctness of the reduction steps, we have to argue that has a d.i.m. if and only if the reduced graph has one (provided that no contradiction arises in the vertex coloring, i.e., it is feasible).

Then let us introduce two reduction steps which will be applied later.

Vertex Reduction. Let . If is forced to be white, then

• color black all neighbors of , and

• remove from .

Let be the reduced subgraph. Clearly, Vertex Reduction is correct, i.e., has a d.i.m. if and only if has a d.i.m.

Edge Reduction. Let . If and are forced to be black, then

• color white all neighbors of and of (other than and ), and

• remove and (and the edges containing or ) from .

Again, clearly, Edge Reduction is correct, i.e., has a d.i.m. if and only if the reduced subgraph has a d.i.m.

The subsequent notions and observations lead to some possible reductions (some of them are mentioned e.g. in [2, 3, 4]).

###### Observation 1 ([2, 3, 4]).

Let be a d.i.m. of .

• contains at least one edge of every odd cycle

in , , and exactly one edge of every odd cycle , , in .

• No edge of any can be in .

• For each either exactly two or none of its edges are in .

Proof. See e.g. Observation 2 in [3].

In what follows, we will also refer to Observation 1  (with respect to ) as to the triangle-property, and to Observation 1  as to the -property.

Since by Observation 1 , every triangle contains exactly one -edge, and the pairwise distance of -edges is at least 2, we have:

###### Corollary 1.

If has a d.i.m. then is -free.

Assumption 1. From now on, by Corollary 1, we assume that the input graph is -free (else it has no d.i.m.).

Clearly, it can be checked (directly) in polynomial time whether the input graph is -free.

By Observation 1 with respect to and the distance property, we have the following:

###### Observation 2.

The mid-edge of any diamond in and the two peripheral edges of any induced butterfly are forced edges of .

Assumption 2. From now on, by Observation 2, we assume that the input graph is (diamond,butterfly)-free.

In particular, we can apply the Edge Reduction to each mid-edge of any induced diamond and to each peripheral edge of any induced butterfly; that can be done in polynomial time.

### 2.3 The distance levels of an M-edge xy in a P3

Based on [4], we first describe some general structure properties for the distance levels of an edge in a d.i.m.  of . Since is , diamond, butterfly)-free, we have:

###### Observation 3.

For every vertex of , is the disjoint union of isolated vertices and at most one edge. Moreover, for every edge , there is at most one common neighbor of and .

Since it is trivial to check whether has a d.i.m.  with exactly one edge, from now on we can assume that . Since is connected and butterfly-free, we have:

###### Observation 4.

If then there is an edge in which is contained in a of .

Proof. Let and assume that is not part of an induced of . Since is connected and , , and since we assume that is not part of an induced of and is - and diamond-free, there is exactly one neighbor of , namely a common neighbor, say of and . Again, since , has a neighbor , and since is - and diamond-free, induce a paw. Clearly, the edge is excluded and has to be dominated by a second -edge, say but now, since is butterfly-free, . Thus, induce a in , and Observation 4 is shown. ∎

Recall [4] for Observation 4. Let be an -edge for which there is a vertex such that induce a with edge . By the assumption that , we have that and are black, and it could lead to a feasible -coloring (if no contradiction arises).

Let and for , let

 Ni(xy):={z∈V:distG(z,xy)=i}

denote the distance levels of . We consider a partition of into , , with respect to the edge (under the assumption that ).

Recall that by (1), is a partition of where is the set of black vertices and is the set of white vertices which is independent.

Since we assume that (and is an edge in a ), clearly, and thus:

 N1 is an independent set of white vertices. (2)

Moreover, no edge between and is in . Since and all neighbors of vertices in are in , we have:

 G[N2] is the disjoint union of edges and isolated vertices. (3)

Let denote the set of edges with and let denote the set of isolated vertices in ; is a partition of . Obviously:

 M2⊆M and S2⊆V(M). (4)

If for , an edge is contained in every d.i.m.  of with , we say that is an -forced -edge, and analogously, if an edge is contained in no d.i.m.  of with , we say that is -excluded. The Edge Reduction for forced edges can also be applied for -forced edges (then, in the unsuccessful case, has no d.i.m. containing ), and correspondingly for -forced white vertices (resulting from the black color of and ), the Vertex Reduction can be applied.

Obviously, by (4), we have:

 Every edge in M2 is an xy-forced M-edge. (5)

Thus, from now on, after applying the Edge Reduction for -edges, we can assume that , i.e., . For every , let denote the -mate of (i.e., ). Let denote the set of -edges with one endpoint in (and the other endpoint in ). Obviously, by (4) and the distance condition for a d.i.m. , the following holds:

 No edge with both ends in N3 and no edge between N3% and N4 is in M. (6)

As a consequence of (6) and the fact that every triangle contains exactly one -edge (recall Observation 1 ), we have:

 For every C3 abc with a∈N3, and b,c∈N4, bc∈M is an xy-forced M-edge. (7)

This means that for the edge , the Edge Reduction can be applied, and from now on, we can assume that there is no such triangle with and , i.e., for every edge in :

 N(u)∩N(v)∩N3=∅. (8)

According to and the assumption that (recall ), let:

1. ,

2. , , and

3. .

By definition, is the set of private neighbors of in (note that ), is a partition of , and is a partition of .

###### Lemma 1 ([4]).

The following statements hold:

1. For all , .

2. For all , is the disjoint union of vertices and at most one edge.

3. is bipartite.

4. , i.e., is an independent subset of white vertices.

5. If a vertex sees two vertices in , , , then is an -forced -edge.

Proof. : Holds by definition of and by the distance condition of a d.i.m. .

: Holds by Observation 3.

: Follows by Observation 1 since every odd cycle in must contain at least one -edge, and by (6).

: If , i.e., sees at least two -vertices then clearly, , and thus, is an independent subset (recall that is an independent set).

: Suppose that sees and in . If then would induce a diamond in . Thus, and now, induce a in ; by Observation 1 , no edge in the is in , and by (6), the only possible -edge for dominating is , i.e., . ∎

By Lemma 1 and the Vertex Reduction for the white vertices of , we can assume:

• , i.e., .

By Lemma 1 , we can assume:

• For , , every vertex has at most one neighbor in .

In particular, if for some , , then there is no d.i.m.  of with , and if , say , then is an -forced -edge. Thus, we can assume:

• For every , .

Let us say that a vertex , , is an out-vertex of if it is adjacent to some vertex of with , or it is adjacent to some vertex of , and is an in-vertex of otherwise.

For finding a d.i.m.  with , one can remove all but one in-vertices; that can be done in polynomial time. In particular, if there is an edge between two in-vertices , , then either or is black, and thus, is completely colored. Thus, let us assume:

• For every , has at most one in-vertex.

###### Lemma 2.

There is no in with vertices , or with vertices , , and , , such that is the center of the .

Proof. First suppose to the contrary that there is an in , say with vertices , , , and center of the . By Observation 5, is the endpoint of a , say with and . But then (with center ) induce an in , which is a contradiction.

Next suppose to the contrary that there is an in , say with vertices , , , such that is the center of the . Again, by Observation 5, is the endpoint of a , say with and . But then (with center ) induce an in , which is a contradiction.

Thus, Lemma 2 is shown. ∎

By Lemma 2, we have:

###### Corollary 2.

The following statements hold:

1. If is part of an edge in then has at most one neighbor in .

2. If is not part of an edge in then has at most two neighbors in .

Proof. : Let for , and suppose to the contrary that has two neighbors , , . Since is diamond-free, and , and since is butterfly-free, . But then (with center ) induce an in , which is a contradiction.

: If has three neighbors , then clearly, is independent (else there would be a triangle and thus, an -edge in – recall (6)), but then it leads again to an in . ∎

###### Lemma 3.

Assume that has a d.i.m.  with . Then there are no three edges between and , , and if there are two edges between and , say and for and then any other vertex in or is white.

Proof. First, suppose to the contrary that there are three edges between and , say , , and for , . Then is black if and only if is white, is black if and only if is white, and is black if and only if is white. Without loss of generality, assume that is black, and is white. Then is white, and is black, but now, and are white, which is a contradiction.

Now, if there are exactly two such edges between and , say , , then again, or is black as well as or is black, and thus, every other vertex in or is white.

Thus Lemma 3 is shown. ∎

By Lemma 3, we can assume:

• For , , there are at most two edges between and .

Then let us introduce the following forcing rules (which are correct). Since no edge in is in (recall (6)), we have:

• All -neighbors of a black vertex in must be colored white, and all -neighbors of a white vertex in must be colored black.

Moreover, we have:

• Every , , should contain exactly one vertex which is black. Thus, if is black then all the remaining vertices in must be colored white.

• If all but one vertices of , , are white and the final vertex is not yet colored, then must be colored black.

Since no edge between and is in (recall (6)), we have:

• For every edge with and , is white if and only if is black and vice versa.

Subsequently, for checking if has a d.i.m.  with , we consider the cases and . In particular, we have the following property:

###### Observation 5.

If for then is an endpoint of an induced , say with vertices such that and with edges , , , , . Analogously, if then is an endpoint of a corresponding induced .

Proof. If then clearly there is such a . Thus, assume that . Then and . Recall that induce a . If then induce a . Thus assume that . Let be a neighbor of . Now, if then induce a , and if but then induce a . Analogously, if then is an endpoint of an induced (which could be part of the above). Thus, Observation 5 is shown. ∎

Let and .

## 3 The Case N4=∅

In this section, we show that for the case (i.e., ), one can check in polynomial time whether has a d.i.m.  with ; we consider the feasible -colorings for . Recall that for every edge , and are black, for , every vertex in is white, and all , , are black, , and recall assumptions (A1)-(A5) and rules (R1)-(R4). In particular, , i.e., .

Clearly, in the case , all the components of can be independently colored. is a trivial component in if has no contact to any other , . Obviously, checking a possible d.i.m.  with can be done easily (and independently) for trivial components; for a vertex let .

From now on we can assume that every component in is nontrivial, i.e., contains at least two , with contact to each other. Every component with at most four -vertices has a polynomial number of feasible -colorings. Thus, we can focus on components with at least five -vertices.

###### Lemma 4.

There is no in with vertices , .

Proof. Suppose to the contrary that there is a in with vertices , . Let be an -neighbor of , and without loss of generality, assume that .

Since (with center ) do not induce an in , we have .

Since (with center ) do not induce an in , we have .

Since (with center ) do not induce an in , we have .

But then (with center ) induce an in , which is a contradiction.

Thus, Lemma 4 is shown. ∎

###### Lemma 5.

If there is a , , , in a nontrivial component of then there is no other edge , , .

Proof. Suppose to the contrary that there is another edge, say for and . Let with . Let with and let be a neighbor of . Clearly, for and for . If and do not have any common neighbor in then there would be an in with center . Thus, assume that . If then (with center ) would induce an . Thus, contacts . Assume without loss of generality that is black. Then is white.

Case 1. is black.

Then are white, and the only possible edges between and are , , or .

If then (with center ) would induce an . Thus, by Lemma 2, .

If and then (with center ) would induce an . Thus, by Lemma 2, either or ; without loss of generality, assume that and . But then