On k-ordered Hamiltonian graphs

  • Authors:

  • H. A. Kierstead;G. N. Sárközy;S. M. Selkow

  • Affiliations:

  • Department of Mathematics, Arizona State University, Tempe, Arizona 85287;Computer Science Department, Worcester Polytechnic Institute, Worcester, Massachusetts 01609;Computer Science Department, Worcester Polytechnic Institute, Worcester, Massachusetts 01609

  • Venue:
  • Journal of Graph Theory
  • Year:
  • 1999

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Abstract

A Hamiltonian graph G of order n is k-ordered, 2 ≤ k ≤ n, if for every sequence v1, v2, …, vk of k distinct vertices of G, there exists a Hamiltonian cycle that encounters v1, v2, …, vk in this order. Define f(k, n) as the smallest integer m for which any graph on n 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 n is sufficiently large in terms of k. Let g(k, n) = $\lceil{{n}\over{2}}\rceil + \lfloor{{k}\over{2}}\rfloor$ - 1. More precisely, we show that f(k, n) = g(k, n) if n ≥ 11k - 3. Furthermore, we show that f(k, n) ≥ g(k, n) for any n ≥ 2k. Finally we show that f(k, n) g(k, n) if 2k ≤ n ≤ 3k - 6. © 1999 John Wiley & Sons, Inc. J Graph Theory 32: 17–25, 1999 An earlier version of this article was written while Sárközy was visiting MSRI Berkeley, as part of the Combinatorics Program.