How large can the domination numbers of a graph be?

A vertex v in a graph G dominates itself as well as its neighbors. A set S of vertices in G is (1) a dominating set if every vertex of G is dominated by some vertex of S, (2) an open dominating set if every vertex of G is dominated by a vertex of S distinct from itself, and (3) a double dominating set if every vertex of G is dominated by at least two distinct vertices of S. The minimum cardinality of a set S satisfying (1), (2), and (3), respectively, is the domination number, open domination number, and double domination number of G, respectively. We consider the problem of determining the maximum value of each of these domination numbers among all graphs of a given order and size. * Research supported in part by the Western Michigan University Faculty Research and Creative Activities Grant Australasian Journal of Combinatorics 21(2000), pp.23-35 1 The Maximum Domination Number of a Graph with Prescribed Order and Size In graph theory we have often been intrigued by how large or how small the value of a graphical parameter can be under various constraints. We discuss three problems of this type, where the parameters involved are various domination numbers and the constraints are given order and size. We refer to books [1, 3] for concepts not defined here. In a graph G a vertex v is said to dominate itself as well as its neighbors. A set S of vertices in G is a dominating set for G if every vertex of G is dominated by some vertex of S. A dominating set of minimum cardinality is a minimum dominating set and its cardinality is the domination number "((G). A comprehensive treatise now exists by Haynes, Hedetniemi and Slater [4] on domination. The first question that we address concerns the largest domination number of a graph of given order n and size m. It is possible to give a complete answer to this question with the significant help of a result of Vizing [5]. Theorem A (Vizing) Let G be a graph of order n and size m. If "((G) 2: 2, then l(n /,(G))(n /,(G) + 2)] m~ 2 . (1) We write max( "(; n, m) for the largest domination number of a graph of order n and size m. Theorem 1.1 For integers n 2: 1 and m with 0 ~ m ~ (~), max(/'; n, m) = In + 1 '1'1 + 2mj . Proof. The result is certainly true for n = 1, so we consider n 2: 2. If m 2: (~) L(n -1)/2J, then every graph G of order n and size m contains a vertex of degree n 1 and so /,( G) = 1. Hence the result holds here as well. Thus it remains to consider a graph G of order n and size m, where n 2: 2 and o ~ m < (~) L (n 1) /2 J. Consequently, max(!'; n, m) 2: 2. Solving for /,( G) in (1), we obtain /,(G) ~ In + 1VI + 2mj, showing that max( /'; n, m) ~ l n + 1 '1'1 + 2m j . To verify the reverse inequality, we construct a ~raph G of order n and size m with /,( G) 2: l n + 1 '1'1 + 2mj = n + 1 r '1'1 + 2m I· Let k = r '1'1 + 2m 1 so that 2 ~ max(!'in,m) ~ n+ 1k; hence k < n. We first construct a graph H according to the parity of k. When k is odd, let M be a perfect matching of Kk+1 and let H = K k+1 M, so /,(H) = 2. For k