14是使得所有立方图的每个4-全着色均为均匀着色的最大阶数

中文摘要 AI 辅助

图的全着色是为其顶点和边分配颜色，使得相邻或关联的元素颜色互不相同；若任意两个颜色类的基数之差不超过1，则称该全着色为均匀着色。Stemock猜想，所有阶数小于20的立方图的每个4-全着色均为均匀着色。本文对该猜想予以否定：环形梯子图L₁₂存在非均匀的4-全着色，且不存在更小的反例——阶数4的情况为空，阶数6、8、10的立方图的每个4-全着色均为均匀着色；同时证明该性质在阶数14时依然成立。我们的证明依赖于一个分解引理：在立方图G的任意4-全着色中，每个颜色类由一个独立集S与G−S的一个完美匹配构成。利用该引理，我们确定了阶数12、16、18时所有可能的颜色类构型，并证明每个列出的构型均可实现。最后，我们给出一个拼接构造，表明对于每个偶数n≥16，存在某个阶数为n的连通立方图，其存在非均匀的4-全着色。综上，14是使得所有立方图的每个4-全着色均为均匀着色的最大阶数。

英文摘要

A total coloring of a graph is an assignment of colors to its vertices and edges so that adjacent or incident elements receive distinct colors, and it is equitable when the cardinalities of any two color classes differ by at most one. Stemock conjectured that every $4$-total coloring of a cubic graph of order less than $20$ is equitable. In this paper, we disprove this conjecture: the circular ladder $L_{12}$ admits a non-equitable $4$-total coloring and, moreover, no smaller counterexample exists: order $4$ is vacuous, and every $4$-total coloring of a cubic graph of order $6$, $8$, or $10$ is equitable. We also prove that the same property holds at order $14$. Our proofs rely on a decomposition lemma, which states that, in any $4$-total coloring of a cubic graph $G$, each color class consists of an independent set $S$ together with a perfect matching of $G-S$. We use the lemma to determine all possible color class configurations for orders $12$, $16$, and $18$, and we show that every listed configuration is attained. Finally, we provide a splicing construction showing that, for every even $n\geq16$, some connected cubic graph of order $n$ admits a non-equitable $4$-total coloring. We may conclude that $14$ is the largest order for which every $4$-total coloring of every cubic graph is equitable.

发表机构

• Universidade Federal do Rio de Janeiro（里约热内卢联邦大学）
• Universidade do Estado do Rio de Janeiro（里约热内卢州立大学）

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