Cycles in Color-Critical Graphs

Cycles in Color-Critical Graphs
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颜色关键图中的循环

DOI:
10.37236/10177
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发表时间:
2019-12
期刊:
The Electronic Journal of Combinatorics
影响因子:
--
通讯作者:
Douglas B. West
Douglas B. West
中科院分区:
其他
文献类型:
--
作者:
Benjamin R. Moore;Douglas B. West

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Tuza[1992]证明了一个没有周期的图的长度等于$1$模$k$的图是$k$可着色的。证明了如果图$G$有一条边$e$使得$G$是$k$-可着色而$G$不可着色,则对于$G$ $,边$e$存在于$G$ $中至少$prod_{i=1}^{r-1} (k-i)$个长度为$1mod r$的$个环,并且$G-e$包含至少$frac12{prod_{i=1}^{r-1} (k-i)}$个长度为$0 mod r$ 0.0d的$个环。$G$的A $(k,d)$-着色是$G$到图$K_{k:d}$的同态,其顶点集${mathbb Z}_{k}$定义为$i$和$j$相邻,如果$dle j-i le k-d$。当$k$和$d$是相对素数时,用$sdequiv 1mod k$定义$s$。Zhu[2002]的结果表明,当$G$没有周期$C$且长度等于$是$k的模$k$时,$G$是$(k,d)$-可着色的。事实上,只需要排除$d$类:我们证明如果$G-e$是$(k,d)$-可着色而$G$不可着色,那么对于${1,ldots,d}$中的某个$i$, $e$至少存在一个长度与$ismod k$相等的循环。此外,如果$iin{1,ldots,d-1}$没有出现这种情况,则$e$至少存在于两个长度为$1mod k$的循环中,并且$G-e$包含一个长度为$0 mod k$的循环。
Tuza [1992] proved that a graph with no cycles of length congruent to $1$ modulo $k$ is $k$-colorable. We prove that if a graph $G$ has an edge $e$ such that $G-e$ is $k$-colorable and $G$ is not, then for $2le rle k$, the edge $e$ lies in at least $prod_{i=1}^{r-1} (k-i)$ cycles of length $1mod r$ in $G$, and $G-e$ contains at least $frac12{prod_{i=1}^{r-1} (k-i)}$ cycles of length $0 mod r$.0D;.A $(k,d)$-coloring of $G$ is a homomorphism from $G$ to the graph $K_{k:d}$ with vertex set ${mathbb Z}_{k}$ defined by making $i$ and $j$ adjacent if $dle j-i le k-d$. When $k$ and $d$ are relatively prime, define $s$ by $sdequiv 1mod k$. A result of Zhu [2002] implies that $G$ is $(k,d)$-colorable when $G$ has no cycle $C$ with length congruent to $is$ modulo $k$ for any $iin {1,ldots,2d-1}$. In fact, only $d$ classes need be excluded: we prove that if $G-e$ is $(k,d)$-colorable and $G$ is not, then $e$ lies in at least one cycle with length congruent to $ismod k$ for some $i$ in ${1,ldots,d}$. Furthermore, if this does not occur with $iin{1,ldots,d-1}$, then $e$ lies in at least two cycles with length $1mod k$ and $G-e$ contains a cycle of length $0 mod k$.
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