### Abstract

A Hamiltonian graph G of order n is k-ordered, 2 ≤ k ≤ n, if for every sequence v_{1}, v_{2}, . . . , v_{k} of k distinct vertices of G, there exists a Hamiltonian cycle that encounters v_{1}, v_{2}, . . . , v_{k} 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) |
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Pages (from-to) | 17-25 |

Number of pages | 9 |

Journal | Journal of Graph Theory |

Volume | 32 |

Issue number | 1 |

DOIs | |

State | Published - Sep 1999 |

### Keywords

- Hamiltonian graph
- k-ordered

### ASJC Scopus subject areas

- Geometry and Topology

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## Cite this

*Journal of Graph Theory*,

*32*(1), 17-25. https://doi.org/10.1002/(sici)1097-0118(199909)32:1<17::aid-jgt2>3.0.co;2-g