Finding dominating induced matchings in S2,2,3-free graphs in polynomial time

doi:10.1016/j.dam.2020.01.028

Discrete Applied Mathematics

Volume 283, 15 September 2020, Pages 417-434

https://doi.org/10.1016/j.dam.2020.01.028 Get rights and content

Abstract

Let $G = (V, E)$ be a finite undirected graph. An edge set $E^{'} \subseteq E$ 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 $NP$ -complete even for very restricted graph classes such as planar bipartite graphs with maximum degree 3 and 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. In this paper, combining two distinct approaches, we solve it in polynomial time for $S_{2, 2, 3}$ -free graphs.

Introduction

Let $G = (V, E)$ be a finite undirected graph. A vertex $v \in V$ dominates itself and its neighbors. A vertex subset $D \subseteq V$ is an efficient dominating set (e.d.s. for short) of $G$ if every vertex of $G$ is dominated by exactly one vertex in $D$ . 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 $G$ (note that not every graph has an e.d.s.)

A set $M$ of edges in a graph $G$ is an efficient edge dominating set (e.e.d.s. for short) of $G$ if and only if it is an e.d.s. in its line graph $L (G)$ . The Efficient Edge Domination (EED) problem asks for the existence of an e.e.d.s. in a given graph $G$ . Thus, the EED problem for a graph $G$ corresponds to the ED problem for its line graph $L (G)$ . 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 $E^{'} \subseteq E$ is a dominating induced matching in $G$ if every edge in $E$ is intersected by exactly one edge of $E^{'}$ .

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 $NP$ -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 $P_{k}$ denote the path with $k$ vertices, say $a_{1}, \dots, a_{k}$ , and $k - 1$ edges $a_{i} a_{i + 1}$ , $1 \leq i \leq k - 1$ . Such a path will also be denoted by $(a_{1}, \dots, a_{k})$ . When speaking about a $P_{k}$ in a graph $G$ , we will always assume that the path is chordless, or, equivalently, that it is an induced subgraph of $G$ .

For indices $i, j, k \geq 0$ , let $S_{i, j, k}$ denote the graph with vertices $u, x_{1}, \dots, x_{i}$ , $y_{1}, \dots, y_{j}$ , $z_{1}, \dots, z_{k}$ such that the subgraph induced by $u, x_{1}, \dots, x_{i}$ forms a $P_{i + 1} (u, x_{1}, \dots, x_{i})$ , the subgraph induced by $u, y_{1}, \dots, y_{j}$ forms a $P_{j + 1} (u, y_{1}, \dots, y_{j})$ , the subgraph induced by $u, z_{1}, \dots, z_{k}$ forms a $P_{k + 1} (u, z_{1}, \dots, z_{k})$ , and there are no other edges in $S_{i, j, k}$ . Vertex $u$ is called the center of this $S_{i, j, k}$ . Thus, claw is $S_{1, 1, 1}$ , and $P_{k}$ is isomorphic to $S_{0, 0, k - 1}$ .

In [10], it is conjectured that for every fixed $i, j, k$ , DIM is solvable in polynomial time for $S_{i, j, k}$ -free graphs (actually, an even stronger conjecture is mentioned in [10]); this includes $P_{k}$ -free graphs for $k \geq 8$ . The following results are known:

Theorem 1

DIM is solvable in polynomial time for

(i)
$S_{1, 1, 1}$ -free graphs [8],
(ii)
$S_{1, 2, 3}$ -free graphs [11],
(iii)
$S_{2, 2, 2}$ -free graphs [10],
(iv)
$S_{1, 2, 4}$ -free graphs [4],
(v)
$S_{1, 1, 5}$ -free graphs [7],
(vi)
$P_{7}$ -free graphs [3] (in this case even in linear time),
(vii)
$P_{8}$ -free graphs [5], and
$(v i i i)$
$P_{9}$ -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 $S_{2, 2, 3}$ -free graphs.

Section snippets

Basic notions

Let $G$ be a finite undirected graph without loops and multiple edges. Let $V$ denote its vertex set and $E$ its edge set; let $n = | V |$ and $m = | E |$ . For $v \in V$ , let $N (v) ≔ {u \in V : u v \in E}$ denote the open neighborhood of $v$ , and let $N [v] ≔ N (v) \cup {v}$ denote the closed neighborhood of $v$ . The degree of $v$ in $G$ is $d_{G} (v) = | N (v) |$ . If $x y \in E$ , we also say that $x$ and $y$ see each other, and if $x y ⁄ \in E$ , we say that $x$ and $y$ miss each other. A vertex set $S$ is independent in $G$ if for every pair of vertices $x, y \in S$ , $x y ⁄ \in E$ . A vertex set $Q$ is a

Coloring $G [X]$

As in [4], we have:

Proposition 1

The following statements hold:

(i)
For every edge $v w \in E$ , $v, w \in N_{3}$ , with $v u_{i} \in E$ and $w u_{j} \in E$ (possibly $i = j$ ), we have $| {v, w} \cap {u_{i}^{'}, u_{j}^{'}} | = 1$ .
(ii)
For every edge $s t \in E$ with $s \in S_{3}$ and $t \in T_{i}$ , $t = u_{i}^{'}$ holds, and thus, $u_{i} t$ is an $x y$ -forced $M$ -edge.

Proof

$(i)$ : By (6), $N_{3}$ does not contain any $M$ -edge, and clearly, if $v w \in E$ then either $v$ or $w$ is black; without loss of generality, let $v$ be black but then $v = u_{i}^{'}$ and $w$ is white, i.e., $w \neq u_{j}^{'}$ .

$(i i)$ : By Observation 12 $(i v)$ , $S_{3} \subseteq I$ and thus, by Proposition 1 $(i)$ , for the edge $s t$

Coloring $G [Y]$

Recall that $X ≔ {x, y} \cup N_{1} \cup N_{2} \cup N_{3}$ and $Y ≔ V ∖ X$ . From now on, let $Y \neq 0̸$ . Subsequently, we apply the polynomial-time solution for $S_{2, 2, 2}$ -free graphs [10] (see Theorem 1 $(i i i)$ ). In particular let us try to connect the “coloring approach” of [10] with the above. Recall that for a d.i.m. $M$ of $G$ , $V (G) = V (M) \cup I$ is a partition of $V (G)$ , all vertices of $V (M)$ are black and all vertices of $I$ are white. Recall the following forcing rules (under the assumption that $x y \in M$ ):

(i)
If a vertex $v$ is white then all of its neighbors

The facilitated case in which $N_{7}$ is empty

Lemma 3

If $N_{7} = 0̸$ then one can detect a white vertex of $H$ in polynomial time.

Proof

Assume $N_{7} = 0̸$ . Then $p \leq 6$ , hence $z \in N_{j}$ with $3 \leq j \leq 5$ . If $z \in N_{5}$ then $V (H) \subseteq N_{6}$ since $N_{7} = 0̸$ , i.e., $d \in N_{6}$ but by Proposition 8, $N_{p - 1} \cap N (d) = 0̸$ , which is a contradiction. Thus, $z \in N_{3} \cup N_{4}$ . Then by the fixed $x y$ -coloring of $G [X]$ (if $z \in N_{3}$ ) and by Proposition 7 (if $z \in N_{4}$ ), the color of $z$ is fixed. Now, since $z, a_{1}, a_{2}$ induce a $C_{3}$ , we have:

$-$
if $z$ is black and $z b_{2} \in E$ , then $b_{2}$ is white;
$-$
if $z$ is black and $z b_{1} \in E$ , then $b_{1}$ is white;
$-$
if $z$ is white, then $d$ is white.

The general case in which $N_{7}$ may be non-empty

By Proposition 10, let us distinguish between $z b_{2} \in E$ and $z b_{1} \in E$ . Then the goal is to detect a white vertex of $S_{2, 2, 2} H$ or a peripheral triangle with a vertex of $H$ . Recall $p ≔ min {i : N_{i} \cap V (H) \neq 0̸}$ .

Proposition 12

If $p \leq 5$ , then one can easily detect a white vertex of $H$ .

Proof

In fact, if $p \leq 5$ , then the color of $z$ (recall $z \in N_{p - 1}$ ) is known by Proposition 7. Then, as in the proof of Lemma 3, one can easily detect a white vertex of $H$ . □

From now on, let us assume that $p \geq 6$ . In particular, let $z_{5} \in N_{p - 5}$ be a neighbor of $z_{4}$ , and let $z_{6} \in N_{p -}$

A polynomial algorithm for DIM on $S_{2, 2, 3}$ -free graphs

The following procedure is part of the algorithm:

Procedure 7.1

DIM-with- $x y$

$(a)$
Set $M ≔ {x y}$ . Determine the distance levels $N_{i} = N_{i} (x y)$ with respect to $x y$ .
$(b)$
Check whether $N_{1}$ is an independent set (see fact (2)) and $N_{2}$ is the disjoint union of edges and isolated vertices (see fact (3)). If not, then STOP with failure.
$(c)$
For the set $M_{2}$ of edges in $N_{2}$ , apply the Edge Reduction for every edge in $M_{2}$ . Moreover, apply the Edge Reduction for each edge $b c$ according to fact ((7)) and then for each edge $u_{i} t_{i}$ according to Observation 12 $(v)$ .
$(d)$
if $N_{4}$

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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Finding dominating induced matchings in S2,2,3-free graphs in polynomial time

Abstract

Introduction

Section snippets

Basic notions

Coloring G[X]

Coloring G[Y]

The facilitated case in which N7 is empty

The general case in which N7 may be non-empty

A polynomial algorithm for DIM on S2,2,3-free graphs

DIM-with-xy

CRediT authorship contribution statement

Acknowledgment

J. Combin. Theory Ser. B

Discrete Appl. Math.

Inform. Process. Lett.

J. Discrete Algorithms

Discrete Appl. Math.

Finding dominating induced matchings in $S_{2, 2, 3}$ -free graphs in polynomial time

Coloring $G [X]$

Coloring $G [Y]$

The facilitated case in which $N_{7}$ is empty

The general case in which $N_{7}$ may be non-empty

A polynomial algorithm for DIM on $S_{2, 2, 3}$ -free graphs

DIM-with- $x y$