On k-ordered Hamiltonian graphs

A Hamiltonian graph G of order a is k-ordered, 2 less than or equal to k less than or equal to n, if for every sequence v(1), v(2),..., v(k), of k distinct vertices of G, there exists a Hamiltonian cycle that encounters vi, v(2),..., v(k) in this order. Define f(k, n) as the smallest integer m for which any graph on a vertices with minimum degree at least m is a k-ordered Hamiltonian graph. In this article, answering a question of Ng and Schultz, we determine f(k, n) if a is sufficiently large in terms of k. Let g(k, n) = [n/2] + [k/2] - 1. More precisely, we show that f(k, n) - g(k, n) if n greater than or equal to 11k - 3 Furthermore, we show that f(k, n) greater than or equal to g(k, n) for any n greater than or equal to 2k. Finally we show that f(k, n) > g(k, n) if 2k less than or equal to n less than or equal to 3k - 6. (C) 1999 John Wiley & Sons, Inc.