论文标题
$ k_6 \ square k_q $的可观数量等于$ 2q+3 $如果$ q \ ge41 $是奇数
The achromatic number of $K_6\square K_q$ equals $2q+3$ if $q\ge41$ is odd
论文作者
论文摘要
令$ g $为图表，$ c $是有限的颜色。顶点着色$ f：v（g）\至c $是完整的，只要对于任何两种不同的颜色$ c_1，c_2 \ in c $ in c $中，就有$ v_1v_2 \ in E（g）$，使得$ f（v_i）= c_i $，$ i = 1,2 $。 $ g $的可观数字是最大数字$ \ mathrm {achr}（g）$ $ g $的完整顶点着色。在本文中，如果$ q \ ge41 $是一个奇数的整数，那么$ k_6 $和$ k_q $的笛卡尔产品的可观数字为$ 2Q+3 $。
Let $G$ be a graph and $C$ a finite set of colours. A vertex colouring $f:V(G)\to C$ is complete provided that for any two distinct colours $c_1,c_2\in C$ there is $v_1v_2\in E(G)$ such that $f(v_i)=c_i$, $i=1,2$. The achromatic number of $G$ is the maximum number $\mathrm{achr}(G)$ of colours in a proper complete vertex colouring of $G$. In the paper it is proved that if $q\ge41$ is an odd integer, then the achromatic number of the Cartesian product of $K_6$ and $K_q$ is $2q+3$.
