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Planar graphs without cycles of length 4 or 5 are $$(2, 0, 0)$$-colorable. (English) Zbl 1327.05073
Summary: Let $$d_1, d_2, \ldots, d_k$$ be $$k$$ nonnegative integers. A graph $$G = (V, E)$$ is $$(d_1, d_2, \ldots, d_k)$$-colorable, if the vertex set $$V$$ of $$G$$ can be partitioned into subsets $$V_1, V_2, \ldots, V_k$$ such that the subgraph $$G [V_i]$$ induced by $$V_i$$ has maximum degree at most $$d_i$$ for $$i = 1, 2, \ldots, k$$. R. Steinberg et al. [Holland. Ann. Discrete Math. 55, 211–248 (1993; Zbl 0791.05044)] conjectured that planar graphs without cycles of length 4 or 5 are $$(0, 0, 0)$$-colorable. O. Hill et al. [Discrete Math. 313, No. 20, 2312–2317 (2013; Zbl 1281.05055)] showed that every planar graph without cycles of length 4 or 5 is $$(3, 0, 0)$$-colorable. In this paper, we show that planar graphs without cycles of length 4 or 5 are $$(2, 0, 0)$$-colorable. For further study in this direction, some problems and conjectures are presented.

##### MSC:
 05C10 Planar graphs; geometric and topological aspects of graph theory 05C15 Coloring of graphs and hypergraphs 05C38 Paths and cycles
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