(Total) Domination in Prisms

Abstract

With the aid of hypergraph transversals it is proved that γ t (Q n+1 ) = 2γ(Q n ), where γ t (G) and γ(G) denote the total domination number and the domination number of G, respectively, and Q n is the n-dimensional hypercube. More generally, it is shown that if G is a bipartite graph, then γ t (G K 2 ) = 2γ(G). Further, we show that the bipartite condition is essential by constructing, for any k ≥ 1, a (non-bipartite) graph G such that γ t (G K 2 ) = 2γ(G) − k. Along the way several domination-type identities for hypercubes are also obtained.

On domination-type invariants of Fibonacci cubes and hypercubes

“…As proved in [32], the equality actually holds here, that is, γ t (Q 2 k +1 ) = 2 2 k −k+1 . More generally, γ t (Q n+1 ) = 2γ(Q n ) holds for any n, a result very recently proved in [6].…”

“…40]. Hence Q 7 and Q 10 are additional sporadic counterexamples (and so are Q 8 and Q 9 since γ t (Q 8 ) = 32 = 2 6 and γ t (Q 9 ) = 64 = 2 7 ).…”

Total domination in generalized prisms and a new domination invariant

“…In Section 2 various properties and bounds for the k-rainbow total domination number are presented. The main theorem of this section shows that for a nontrivial graph G of order n we have γ krt (G) = n as soon as k ≥ 2∆(G), where ∆(G) denotes the maximum degree of G. In Section 3 we complement a result by Azarija et al [3], who considered the total domination of the Cartesian products C n 2 K 2 when n = 6ℓ+1 for ℓ ≥ 1, or n is an even number, by establishing a complete formula for γ krt (C n ) = γ t (C n 2 K k ) for arbitrary k. In addition, exact values of the k-rainbow total domination number of paths for every k are given. Along the way it is demonstrated how a new weight-redistribution method can be used to find a lower bound for a domination invariant.…”

“…Brešar et al [4] extended their result by characterizing the pairs of graphs G and H for which 2γ t (G 2 H) = γ t (G)γ t (H), whenever γ t (H) = 2. A recent result of Azarija et al [3] that γ t (Q n+1 ) = 2γ(Q n ) holds for all n ≥ 1 follows from a more general result on bipartite prisms.…”

“…Thus, if k = 2 and G is either a path or an even cycle, by Observation 3 and Corollary 9 we immediately obtain γ 2rt (G) = 2γ(G) = 2 n 3 . Azarija et al [3] also considered cycles C n where n = 6ℓ + 1 and ℓ ≥ 1, and obtained that in this case γ 2rt (C n ) = 2 n 3 − 1. To settle the remaining cases, i.e., n = 6ℓ + 3 and n = 6ℓ + 5, recall the Theorem 2.11 from [15], which states that for a graph G of order n with no isolated vertex it holds γ t (G) ≥ n ∆(G) .…”