The binding number of a random graph

Let G be a random graph with n labelled vertices in which the edges are chosen independently with a fixed probability p, 0 < p < 1. In this note we prove that, with the probability tending to 1 as n -+ 00, the binding number of a random graph G satisfies: (i) b(G) = (n l)/(n 8), where 8 is the minimal degree of G; (ii) l/q E < b(G) < l/q, where E is any fixed positive number and q = 1 p; (iii) b(G) is realized on a unique set X = V( G)\N(x), where deg(x) = 8(G), and the induced subgraph (X) contains exactly one isolated vertex x. All graphs will be finite and undirected, without loops or multiple edges. If G is a graph, V(G) denotes the set of vertices in G, and n =1 V(G) I. We shall denote the neighborhood of a vertex x by N(x). More generally, N(X) UxEX N(x) for X ~ V( G). The minimal degree of vertices and the vertex connectivity of G are denoted by 8 = 8( G) and K,( G), respectively. For a set X of vertices, (X) deno tes the subgraph of G induced by X. Woodall [5] defined the binding number b( G) of a graph G as follows: b(G) = . I N(X) I Tl~ IX I ' where :F = {X : 0 =IX ~ V(G), N(X) =IV(G)}. We say that b(G) is realized on a set X if X E :F and b( G) N (X) I / 1 X I, and the set X is called a realizing set for b( G). Proposition 1 For any graph G) _8 _ ::; b( G) ::; n 1 n-8 n*On leave from Faculty of Mechanics and Mathematics, Belarus State Uni versity, Minsk 220050, Republic of Belarus. Supported by an award from the Alexander von Hu mboldt Foundation. Australasian Journal of Combinatorics 15(1997), pp.271-275 Proof. The upper bound is proved by Woodall in [5]. Let us prove the lower bound. Let X E :F and I N(X) I j I X 1= b(G), i.e., X is a realizing set. We have I N(X) 12:: 0, since the set X is not empty. Suppose that I X 12:: n 0 + 1. Then any vertex of G is adjacent to some vertex of X, i.e. N(X) = V(G), a contradiction. Therefore I X I:::; n 0 and b(G) =1 N(X) I j I X 12:: oj(n 0). The proof is complete. III Note that the difference between the upper and lower bounds on b( G) in Proposition 1 is less than 1. In the sequel we shall see that the binding number of almost every graph is equal to the upper bound in Proposition 1. Let 0 < p < 1 be fixed and put q = 1 p. Denote by Q(n,P(edge) = p) the discrete probability space consisting of all graphs with n fixed and labelled vertices, in which the probability of each graph with M edges is pM qN-M, where N = (~). Equivalently, the edges of a labelled random graph are chosen independently and with the same probability p. We say that a random graph G satisfies a property Q if P(G has Q) -+ 1 as n -+ 00. We shall need the following results. Theorem 1 (Bollobas [1]) A random graph G satisfies h;(G) = J(G). Theorem 2 (Bollobas [1]) A random graph G satisfies I o(G) pn + (2pqn log n)1/2 ( Plqn ) 1/2 log log n I :::; C(n) (-1 n ) 1/2, 80gn ogn where C(n) -+ 00 arbitrarily slowly. Theorem 3 (Erdos and Wilson [3]) A random graph has a unique vertex of minimal degree. Now we can state the main result of the paper. Theorem 4 The binding number of a random graph G satisfies b(G) = n 1 nProof. Taking into account Proposition 1, it is sufficient to prove that for any set X E F. Let Y = N(X) \X and consider three cases. (i) The induced subgraph (X) does not contain an isolated vertex. The set V(G)\N(X) is not empty, since X E F. Hence the set Y is a cutset of the graph G. By Theorem 1, h;(G) = o(G). -Therefore I Y 12:: 0 and I X 1< n o. We have I N(X) I = I Y I + I X I = ~ + 1 > _n_ > ~. IXI IXI IXI -n 0 n-O