Recognition: unknown
A relaxation of the strong Bordeaux Conjecture
classification
🧮 math.CO
keywords
colorablegraphmathcalcdotsldotsadjacentbordeauxborodin
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Let $c_1, c_2, \cdots, c_k$ be $k$ non-negative integers. A graph $G$ is $(c_1, c_2, \cdots, c_k)$-colorable if the vertex set can be partitioned into $k$ sets $V_1,V_2, \ldots, V_k$, such that the subgraph $G[V_i]$, induced by $V_i$, has maximum degree at most $c_i$ for $i=1, 2, \ldots, k$. Let $\mathcal{F}$ denote the family of plane graphs with neither adjacent 3-cycles nor $5$-cycle. Borodin and Raspaud (2003) conjectured that each graph in $\mathcal{F}$ is $(0,0,0)$-colorable. In this paper, we prove that each graph in $\mathcal{F}$ is $(1, 1, 0)$-colorable, which improves the results by Xu (2009) and Liu-Li-Yu (2014+).
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