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) = [n/2] + [k/2] - 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.
|Original language||English (US)|
|Number of pages||9|
|Journal||Journal of Graph Theory|
|State||Published - Sep 1999|
- Hamiltonian graph
ASJC Scopus subject areas
- Geometry and Topology