The Domination number of (Kp, P5)-free graphs

We prove that, for each p ≥ 1, there exists a polynomial time algorithm for finding a minimum domination set in the class of all (Kp, P5)-free graphs. Let G be a graph with vertex-set V (G) and edge-set E(G). The notation x ∼ y (respectively, x 6∼ y) means that vertices x, y ∈ V (G) are adjacent (respectively, non-adjacent). Moreover, if X ⊆ V (G) and y ∈ V (G)\X, we write y ∼ X (respectively, y 6∼ X) to indicate that y is adjacent (respectively, non-adjacent) to all vertices in X. The neighborhood of a vertex x ∈ V (G) is the set N(x) = NG(x) = {y ∈ V (G) : x ∼ y}; the closed neighborhood of x is N [x] = {x} ∪ N(x). Similarly, for a set X ⊆ V (G), N(X) = ⋃ x∈X N(x) and N [X] = X ∪ N(X). We use the notation Pn and Kn for a path and a complete graph of order n ≥ 1, respectively. A set D ⊆ V (G) is a domination set in a graph G if every vertex of V (G)\D is adjacent to a vertex of D. The domination number γ(G) of a graph G is the minimum cardinality of a dominating set in G. A dominating set G in G is minimum if |D| = γ(G). For a set X ⊆ V (G) we say that X dominates N [X]. Let Z be a set of graphs. A graph G is called Z-free if G does not contain any graph of Z as an induced subgraph. It is well known (see Bertossi [1], Johnson [3], and Korobitsin [4]) that the problem of finding a minimum dominating set is NP-complete for both P5-free graphs and Kp-free graphs (p ≥ 3). We prove that this problem can be solved in polynomial time for (Kp, P5)-free graphs. Definition 1 For n ≥ m ≥ 1 we define a graph H = S(n,m) as follows: V (H) = A ∪ B, where A = {u1, u2, . . . , un} and B = {v1, v2, . . . , vm} are disjoint sets, and E(H) = {uiuj : i, j ∈ {1, 2, . . . , n}, i 6= j} ∪ {uivi : i = 1, 2, . . . ,m}. Any graph S(n,m) (n and m are not fixed) will be called a simple split graph (Figure 1). All graph of the form S(n,m) are split graphs in sense of Földes and Hammer [2]. uu uu uu u u r r r r r r r r r A B complete subgraph u1 u2 um um+1 un