Proof. Let G satisfy the hypothesis of Theorem 1. Clearly, G contains a circuit ; let C be the longest one . If G has no Hamiltonian circuit, there is a vertex x with x ~ C . Since G is s-connected, there are s paths starting at x and terminating in C which are pairwise disjoint apart from x and share with C just their terminal vertices x l, X2, . . ., x s (see [ 11, Theorem 1) . For each i = 1, 2, . . ., s, let y i be the successor of x i in a