🤖 AI Summary
本文通过分解引理等方法，解决了Stemock关于4-全染色在立方图中是否总是公平的猜想，证明了14阶是所有立方图4-全染色都公平的最大阶数。
📝 Abstract
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.