English

Cycles in Color-Critical Graphs

Combinatorics 2021-11-16 v4

Abstract

Tuza [1992] proved that a graph with no cycles of length congruent to 11 modulo kk is kk-colorable. We prove that if a graph GG has an edge ee such that GeG-e is kk-colorable and GG is not, then for 2rk2\leq r\leq k, the edge ee lies in at least i=1r1(ki)\prod_{i=1}^{r-1}(k-i) cycles of length 1modr1\mod r in GG, and GeG-e contains at least 12i=1r1(ki)\frac{1}{2}\prod_{i=1}^{r-1}(k-i) cycles of length 0modr0 \mod r. A (k,d)(k,d)-coloring of GG is a homomorphism from GG to the graph Kk:dK_{k:d} with vertex set Zk\mathbb{Z}_{k} defined by making ii and jj adjacent if djikdd\leq j-i \leq k-d. When kk and dd are relatively prime, define ss by sd1modksd\equiv 1\mod k. A result of Zhu [2002] implies that GG is (k,d)(k,d)-colorable when GG has no cycle CC with length congruent to isis modulo kk for any i{1,,2d1}i\in \{1,\ldots,2d-1\}. In fact, only dd classes need be excluded: we prove that if GeG-e is (k,d)(k,d)-colorable and GG is not, then ee lies in at least one cycle with length congruent to ismodkis\mod k for some ii in {1,,d}\{1,\ldots,d\}. Furthermore, if this does not occur with i{1,,d1}i\in\{1,\ldots,d-1\}, then ee lies in at least two cycles with length 1modk1\mod k and GeG-e contains a cycle of length 0modk0 \mod k.

Keywords

Cite

@article{arxiv.1912.03754,
  title  = {Cycles in Color-Critical Graphs},
  author = {Benjamin Moore and Douglas B. West},
  journal= {arXiv preprint arXiv:1912.03754},
  year   = {2021}
}

Comments

10 pages

R2 v1 2026-06-23T12:39:25.971Z