On the equality of domination number and 2-domination number

Abstract

The 2-domination number γ 2 (G) of a graph G is the minimum cardinality of a set D ⊆ V (G) for which every vertex outside D is adjacent to at least two vertices in D. Clearly, γ 2 (G) cannot be smaller than the domination number γ(G). We consider a large class of graphs and characterize those members which satisfy γ 2 = γ. For the general case, we prove that it is NP-hard to decide whether γ 2 = γ holds. We also give a necessary and sufficient condition for a graph to satisfy the equality hereditarily.

Bipartite graphs with close domination and k-domination numbers

“…We also proved that it is NP-hard to decide whether this equality holds for a graph. Moreover, we gave a necessary and sufficient condition for a graph to satisfy γ 2 (G) = γ(G) hereditarily [8]. Some similar problems involving different domination-type graph and hypergraph invariants were considered for example in [1,2,7,18,21,23].…”

“…In particular, for each k-uniform hypergraph F , we have exactly one graph in B * k having F as its underlying hypergraph, but there are infinitely many such graphs in B k . Note that in [8], where (γ, γ 2 )graphs were studied, a graph class G was introduced that exactly corresponds to the class G 2 defined here. Moreover, by the results in [1,18,22], the class B 2 is the collection of those connected graphs which satisfy τ (G) = γ(G) and δ(G) ≥ 2.…”

“…In order to prove that the corresponding decision problem is NP-hard, we construct a polynomial time reduction from 3-SAT problem, a classical NP-complete problem [13]. Note that we adopt the approach we used in [8] to prove NP-hardness. Since the construction we need here is more general, we only present the construction explicitly and give a sketch of the rest of the proof, for the brevity.…”

“…As proved in [8], S 2 is the set of all (γ, γ 2 )-perfect graphs. Since each S k contains only bipartite graphs, it is equivalent to saying that S 2 is the set of all bipartite (γ, γ 2 )-perfect graphs.…”

“…, v k−1 0 }∩ D ′ = ∅. There are three cases to consider and for a detailed discussion, we refer the reader to[8].…”