Cycles in color-critical graphs

Benjamin R. Moore¹ ; Douglas B. West²

¹Charles University
²University of Illinois - Urbana

The electronic journal of combinatorics, Tome 28 (2021) no. 4

Résumé

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 ${\mathbb Z}_{k}$ 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 $0 \mod k$. An addendum was added on December 13, 2021.

Détail

DOI : 10.37236/10177

Classification : 05C15, 05C38, 05C12
Mots-clés : Minty's theorem, generalized Kempe chain technique