### Cycles in Color-Critical Graphs

Benjamin R. Moore, Douglas B. West
2021 Electronic Journal of Combinatorics
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 $2\le r\le k$, the edge $e$ lies in at least $\prod_{i=1}^{r-1} (k-i)$ cycles of length $1\mod r$ in $G$, and $G-e$ contains at least $\frac12{\prod_{i=1}^{r-1} (k-i)}$ cycles of length $0 \mod r$. A $(k,d)$-coloring of $G$ is a homomorphism from $G$ to the graph $K_{k:d}$ with vertex set
more » ... athbb Z}_{k}$defined by making$i$and$j$adjacent if$d\le j-i \le k-d$. When$k$and$d$are relatively prime, define$s$by$sd\equiv 1\mod 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$for any$i\in \{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$lies in at least one cycle with length congruent to$is\mod k$for some$i$in$\{1,\ldots,d\}$. Furthermore, if this does not occur with$i\in\{1,\ldots,d-1\}$, then$e$lies in at least two cycles with length$1\mod k$and$G-e$contains a cycle of length$0 \mod k\$.