Vertices in all minimum paired-dominating sets of block graphs

Let $G=(V,E)$ be a simple graph without isolated vertices. A set $S\subseteq V$ is a paired-dominating set if every vertex in $V-S$ has at least one neighbor in $S$ and the subgraph induced by $S$ contains a perfect matching. In this paper, we present a linear-time algorithm to determine whether a given vertex in a block graph is contained in all its minimum paired-dominating sets

The Paired Domination Number of Cubic Graphs

Let G be a simple undirected graph with no isolated vertex. A paired dominating set of G is a dominating set which induces a subgraph that has a perfect matching. The paired domination number of G, denoted by {\gamma}pr(G), is the size of its smallest paired dominating set. Goddard and Henning conjectured that {\gamma}pr(G) {\leq} 4n/7 holds for every graph G with {\delta}(G) {\geq} 3, except the Petersen Graph. In this paper, we prove this conjecture for cubic graphs

An O(n)-time Algorithm for the Paired-Domination Problem on Permutation Graphs

A vertex subset D of a graph G is a dominating set if every vertex of G is either in D or is adjacent to a vertex in D. The paired-domination problem on G asks for a minimum-cardinality dominating set S of G such that the subgraph induced by S contains a perfect matching; motivation for this problem comes from the interest in finding a small number of locations to place pairs of mutually visible guards so that the entire set of guards monitors a given area. The paired-domination problem on general graphs is known to be NP-complete. In this paper, we consider the paired-domination problem on permutation graphs. We define an embedding of permutation graphs in the plane which enables us to obtain an equivalent version of the problem involving points in the plane, and we describe a sweeping algorithm for this problem; given the permutation over the set Nn = {1, 2,..., n} defining a permutation graph on n vertices, our algorithm computes a paired-dominating set of the graph in O(n) time, and is therefore optimal