AI 中文总结
本文证明图的配对哈密顿性质在笛卡尔积下保持，并给出充要条件，指出无支撑限制时逆命题不成立。
AI 中文摘要
设 $G$ 为阶数至少为 4 的偶阶简单图，并设 $K_G$ 表示 $V(G)$ 上的完全图。$K_G$ 的完美匹配称为 $G$ 的配对。若 $G$ 的每个配对 $M$ 都允许存在一个与 $M$ 不相交的完美匹配 $N\subseteq E(G)$，使得 $M\cup N$ 是 $K_G$ 的哈密顿圈，则称图 $G$ 具有配对哈密顿性质（PH-性质）。我们证明 PH-性质在笛卡尔积下保持。更精确地，对于阶数至少为 4 的偶阶图 $G$ 和 $H$，我们证明 $G$ 和 $H$ 均为 PH 当且仅当 $G\square H$ 的每个配对都允许一个哈密顿补全，该补全包含在由 $G$ 或 $H$ 的完美匹配所确定的顶点不相交棱柱的生成并中。若无此支撑限制，逆命题不成立：即使两个图均非 PH，其笛卡尔积也可能是 PH。
英文摘要
Let $G$ be a simple graph of even order at least four, and let $K_G$ denote the complete graph on $V(G)$. A perfect matching of $K_G$ is called a pairing of $G$. The graph $G$ has the Pairing-Hamiltonian property, or PH-property, if every pairing $M$ of $G$ admits a perfect matching $N\subseteq E(G)$, disjoint from $M$, such that $M\cup N$ is a Hamiltonian cycle of $K_G$. We prove that the PH-property is preserved under Cartesian products. More precisely, for graphs $G$ and $H$ of even order at least four, we show that both $G$ and $H$ are PH if and only if every pairing of $G\square H$ admits a Hamiltonian completion contained in a spanning union of vertex-disjoint prisms determined by a perfect matching of $G$ or of $H$. Without this support restriction, the converse fails: a Cartesian product may be PH even when neither of the two graphs is PH.
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