On the total chromatic number of the direct product of cycles and complete graphs

Diane Castonguay, Celina de Figueiredo, Luis Kowada, C. Patrão, Diana Sasaki, Mario Valencia-Pabon
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Abstract

A \textit{$k$-total coloring} of a graph $G$ is an assignment of $k$ colors to the elements (vertices and edges) of $G$ so that adjacent or incident elements have different colors. The total chromatic number is the smallest integer $k$ for which $G$ has a $k$-total coloring. The well known Total Coloring Conjecture states that the total chromatic number of a graph is either $\Delta(G)+1$ (called Type~1) or $\Delta(G)+2$ (called Type~2), where $\Delta(G)$ is the maximum degree of $G$.  We consider the direct product of complete graphs $K_m \times K_n$.  It is known that if at least one of the numbers $m$ or $n$ is even, then  $K_m \times K_n$ is Type~1, except for $K_2 \timesK_2$.  We prove that the graph $K_m \times K_n$ is Type~1 when both $m$ and $n$ are odd numbers, by using that the conformable condition is sufficient for the graph $K_m \times K_n$ to be Type~1 when both $m$ and $n$ are large enough, and by constructing the target total colorings by using Hamiltonian decompositions and a specific color class, called guiding color.  We additionally apply our technique to the direct product $C_m \times K_n$ of a cycle with a complete graph. Interestingly, we are able to find a Type 2 infinite family $C_m \times K_n$, when $m$ is not a multiple of 3 and $n = 2$. We provide evidence to conjecture that all other $C_m \times K_n$ are Type 1.
论循环与完整图直积的总色度数
图 $G$ 的总着色(textit{$k$-total coloring})是将 $k$ 颜色分配给图 $G$ 的元素(顶点和边),使相邻或相交的元素具有不同的颜色。总着色数是 $G$ 具有 $k$ 总着色的最小整数 $k$。众所周知的总着色猜想指出,图的总色度数要么是 $\Delta(G)+1$(称为 Type~1),要么是 $\Delta(G)+2$(称为 Type~2),其中 $\Delta(G)$ 是 $G$ 的最大度数。我们考虑完整图 $K_m \times K_n$ 的直接乘积。众所周知,如果 $m$ 或 $n$ 中至少有一个数是偶数,那么 $K_m \times K_n$ 就是 Type~1,只有 $K_2 \timesK_2$ 除外。我们证明了当 $m$ 和 $n$ 都是奇数时,图 $K_m \times K_n$ 是 Type~1,方法是利用符合条件足以使图 $K_m \times K_n$ 在 $m$ 和 $n$ 都足够大时是 Type~1,并通过使用哈密尔顿分解和一个特定的颜色类(称为引导色)来构造目标总着色。有趣的是,当 $m$ 不是 3 的倍数且 $n = 2$ 时,我们能够找到一个类型 2 的无穷族 $C_m \times K_n$.我们提供了证据来猜想所有其他的 $C_m \times K_n$ 都是类型 1。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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