Petersen coloring conjecture

Importance: High ✭✭✭
Author(s): Jaeger, Francois
Subject: Graph Theory
» Coloring
» » Edge coloring
Recomm. for undergrads: no
Prize: none
Posted by: mdevos
on: March 7th, 2007

\begin{conjecture} Let $G$ be a \Def[cubic]{cubic graph} graph with no \Def[bridge]{bridge (graph theory)}. Then there is a coloring of the edges of $G$ using the edges of the \Def[Petersen]{petersen graph} graph so that any three mutually adjacent edges of $G$ map to three mutually adjancent edges in the Petersen graph. \end{conjecture}

This extrordainary conjecture asserts that in a very strong sense, every bridgeless cubic graph has all of the cycle-space properties posessed by the Petersen graph. If true, this conjecture would imply both \OPref[The Berge-Fulkerson conjecture]{the_berge_fulkerson_conjecture} and \OPref[The five cycle double cover conjecture]{m_n_cycle_covers}.

If $G$ is a graph and $C \subseteq E(G)$ we say that $C$ is a binary cycle if every vertex in the graph $(V(G),C)$ has even degree. If $H$ is a graph and $f : E(G) \rightarrow E(H)$ is a map, we say that $f$ is cycle-continuous if the pre-image of every binary cycle is a binary cycle. The following conjecture is an equivalent reformulation of the Petersen coloring conjecture.

\begin{conjecture}[Petersen coloring conjecture (2)] Every bridgeless graph has a cycle-continuous mapping to the Petersen graph. \end{conjecture}

Question

For which bridgeless cubic graphs has this been checked for?

Re:

It is trivially true for all that are 3-edge-colorable -- which is the vast majority. Among the rest, I checked it using computer and lists of snarks for all graphs upto 34 vertices. (And some more -- e.g. all flower-snarks.)

Best wishes, Robert

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