Non-conflicting no-where zero $Z_2\times Z_2$ flows in cubic graphs

TL;DR

This paper explores non-conflicting no-where zero flows in cubic graphs and their implications for 6-edge-colorings, establishing new connections with graph structure and conjectures on perfect matchings.

Contribution

It introduces the concept of non-conflicting flows in cubic graphs and demonstrates their role in ensuring normal 6-edge-colorings, extending understanding of graph colorings and flows.

Findings

Non-conflicting flows imply the existence of normal 6-edge-colorings in certain cubic graphs.

Claw-free and specific 2-factor cubic graphs admit such 6-edge-colorings.

Constructed examples show limitations of non-conflicting flows in some 2-edge-connected cubic graphs.

Abstract

Let $Z_2\times Z_2=\{0, \alpha, \beta, \alpha+\beta\}$. If $G$ is a bridgeless cubic graph, $F$ is a perfect matching of $G$ and $\overline{F}$ is the complementary 2-factor of $F$, then a no-where zero $Z_2\times Z_2$-flow $\theta$ of $G/\overline{F}$ is called non-conflicting with respect to $\overline{F}$, if $\overline{F}$ contains no edge $e=uv$, such that $u$ is incident to an edge with $\theta$-value $\alpha$ and $v$ is incident to an edge with $\theta$-value $\beta$. In this paper, we demonstrate the usefulness of non-conflicting flows by showing that if a cubic graph $G$ admits such a flow with respect to some perfect matching $F$, then $G$ admits a normal 6-edge-coloring. We use this observation in order to show that claw-free bridgeless cubic graphs, bridgeless cubic graphs possessing a 2-factor having at most two cycles admit a normal 6-edge-coloring. We demonstrate the usefulness of non-conflicting flows further by relating them to a recent conjecture of Thomassen about edge-disjoint perfect matchings in highly connected regular graphs. In the end of the paper, we construct infinitely many 2-edge-connected cubic graphs such that $G/\overline{F}$ does not admit a non-conflicting no-where zero $Z_2\times Z_2$-flow with respect to any perfect matching $F$.