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arxiv: math/9612212 · v1 · pith:O6WBB7MEnew · submitted 1996-12-04 · 🧮 math.CO

On k-ordered Hamiltonian Graphs

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keywords graphhamiltonianorderedfracorderbounddegreeevery
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A Hamiltonian graph $G$ of order $n$ is $k$-ordered, $2\leq k \leq n$, if for every sequence $v_1, v_2, \ldots ,v_k$ of $k$ distinct vertices of $G$, there exists a Hamiltonian cycle that encounters $v_1, v_2, \ldots , v_k$ in this order. In this paper, answering a question of Ng and Schultz, we give a sharp bound for the minimum degree guaranteeing that a graph is a $k$-ordered Hamiltonian graph under some mild restrictions. More precisely, we show that there are $\varepsilon, n_0> 0$ such that if $G$ is a graph of order $n\geq n_0$ with minimum degree at least $\lceil \frac{n}{2} \rceil + \lfloor \frac{k}{2} \rfloor - 1$ and $2\leq k \leq \eps n$, then $G$ is a $k$-ordered Hamiltonian graph. It is also shown that this bound is sharp for every $2\leq k \leq \lfloor \frac{n}{2} \rfloor$.

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