A sufficient condition for Hamiltonian cycles in bipartite tournaments

We prove a new sufficient conditi()n on degrees for a bipartite tournament to be Hamiltonian, that is, if an n x n bipartite tournament T satisfies the condition dT(u) + dj;(v) ~ n 1 whenever uv is an arc of T, then T is Hamiltonian, except for two exceptional graphs. This result is shown to be best possible in a sense. T(X, Y, E) denotes a bipartite tournament with bipartition (X, Y) and vertex-set VeT) = XuY and arc-set E(T). If IXI == m and IYI = n, such a bipartite tournament is called an m x n bipartite tournament. For a vertex v of T and a sub digraph S of T, we define N,I-(v) and N:(v) to be the set of vertices of S which, respectively, dominate and are dominated by, the vertex v. Put Ni(S) = UNi(v)i Ni(S) = UNi(v)j vEil vEil dT(v) = I Ni(v) I j dj:(v) = I N:i(v) I· Let P be a subset of X and Q a subset of Yj P --+ Q (resp. Q --. P) denotes pq E E(T) (resp. qp E E(T)) for all pEP and all q E Q. If P = {xl this becomes x --+ Q. To simplify notation, we denote also B1 --+ B2, B2 --+ Ba,' ", by B1 --+ B2 --+ Ba --+ .. '. Moreover, a fa.ctor of T is a spanning sub digraph H of T such that dir( v) = dk( v) = 1 for all v E VeT). T is said to be strong if for any two vertices u and v, there is a path from u to v and a path from v to u. A component of T is a maximal strong subdigraph. T(bt, b2 , ba, b4 ) denotes the bipartite tournament, whose vertex-set may be partitioned into four independent sets Bi , i = 1,2,3,4, such tha.t IBil = bi .. ~. 0 and B1 --+ B2 --. Ba --+ B4 --+ B1. Other terms and symbols not defined in this paper can be found in [1]. Up to now, there are very few conditions that imply the existence of Hamiltonian cycles for bipartite tournaments. An obvious necessary condition for an m X n bipartite tournament to be Hamiltonian is m = n. Therefore, we are only interested in researching Hamiltonian properties in n x n bipartite tournaments. We recall now the well-known conditions for an n x n bipartite tournament to have Hamiltonian cycles. lThe work for this paper was supported by Youth Science Foundation of Taiyuan Institute of Machinery and Natural Science Foundation of Shanxi Province Australasian Journal of Combinatorics 5( 1992 L pp.299-304 Theorem 1 (Jackson (2}). If an n x n strong bipartite tournament T satisfies vu tfE(T) => dT(u) +df(v) 2:: n, then T is Hamiltonian. Theorem 2 (Wang [3}). Ifan n x n bipartite tournament T satisfies vu tfE(T) => dT(u) + df(v) 2:: n 1, then T is Hamiltonian, unless n is odd and T is isomorphic to T(~,~, n;l ) n;l). Obviously, Theorem 2 improves Theorem 1. In this paper we prove another condition that is weaker than the conditions of the two theorems above, ensuring an n x n bipartite tournament to be Hamiltonian, except for two described cases. In showing the main result we will use the following theorem: Theorem 3 (Haggkvist and Manoussakis [4}). A bipartite tournament T is Hamiltonian if and only if T is strong and contains a factor. Theorem 4 If an n x n bipartite tournament T satisfies uv E E(T) => dT(u) + df(v) 2:: n 1, then T is Hamiltonian, unless T is isomorphic to T(ntl,~, n;l, n;l) when n is odd T( n n-2 n!!.±l) h . or 2' -2-' 2' 2 W en n ~s even. Proof. Suppose that T is an n x n bipartite tournament satisfying the hypotheses of the theorem. We first establish two claims. Claim 1. If n 2:: 3, then T is strong. Assume that T is not strong and has components B l , B2 ,"', Bm with m 2:: 2 such that X(Bi) ~ Y(Bj ) and Y(Bi) ~ X(Bj) whenever i ~ j. Then Bl contains a vertex u such that dT(u) ~ !V<:l)l. Such a vertex exists because Without loss of generality we may assume that u E X. . Case 1. Y(Bm) -# 0. If there is a vertex v in Y(Bm) such that df(v) ~ !V(!",)!, then we have In particular, uv E E(T) implies dT(u) + 4(v) 2:: n-1