Cycle-Pancyclism in Tournaments III

Let T be a hamiltonian tournament with n vertices and fl a hamiltonian cycle of T . For a cycle C k of length k in T we denote I fl (C k ) = jA(fl) " A(C k )j, the number of arcs that fl and C k have in common. Let f(n; k; T; fl) = maxfI fl (C k )jC k ae Tg and f(n; k) = minff(n; k; T; fl)jT is a hamiltonian tournament with n vertices, and fl a hamiltonian cycle of Tg. In a previous paper [3] we studied the case of n 2k \Gamma 4 and proved that ffl f(n; 3) = 1, f(n; 4) = 1 and f(n; 5) = 2 if n 6= 2k \Gamma 2; ffl f(n; k) = k \Gamma 1 if and only if n = 2k \Gamma 2; ffl for k ? 5, f(n; k) = k \Gamma 2 if and only if n 2k \Gamma 4, n 6= 2k \Gamma 2 and n j k (mod k \Gamma 2); ffl for k ? 5, f(n; k) = k \Gamma 3 if and only if n 2k \Gamma 4 and n 6j k (mod k \Gamma 2). In this paper we consider the case of n 2k \Gamma 5 and complete the description of f(n; k) by proving that f(n; k) = k \Gamma 4 if and only if n 2k \Gamma 5. 1 Introduction The subject of pancyclism in tournament..