Planar graphs without cycles of length 4 or 5 are \((2, 0, 0)\)-colorable

DOI10.1016/J.DISC.2015.10.029zbMath1327.05073OpenAlexW2150902778MaRDI QIDQ898156

Jinghan Xu, Ming Chen, Pei-Pei Liu, Ying Qian Wang

Publication date: 8 December 2015

Published in: Discrete Mathematics (Search for Journal in Brave)

Full work available at URL: https://doi.org/10.1016/j.disc.2015.10.029

zbMATH Keywords

bad cyclesdischargingimproper coloringSteinberg's conjecturesuper extension

Mathematics Subject Classification ID

Paths and cycles (05C38) Planar graphs; geometric and topological aspects of graph theory (05C10) Coloring of graphs and hypergraphs (05C15)

Related Items (15)

Every planar graph without cycles of length 4 or 9 is \((1, 1, 0)\)-colorable ⋮ Every planar graph with girth at least 5 is \((1,9)\)-colorable ⋮ New restrictions on defective coloring with applications to Steinberg-type graphs ⋮ Every planar graph without 4-cycles and 5-cycles is (3,3)-colorable ⋮ Planar graphs without 5-cycles and intersecting triangles are \((1, 1, 0)\)-colorable ⋮ The \((3, 3)\)-colorability of planar graphs without 4-cycles and 5-cycles ⋮ Every planar graph without 4-cycles and 5-cycles is \((2, 6)\)-colorable ⋮ Planar graphs without adjacent cycles of length at most five are (2, 0, 0)-colorable ⋮ Decomposing a planar graph without triangular 4-cycles into a matching and a 3-colorable graph ⋮ Planar graphs without 4-cycles and close triangles are \((2,0,0)\)-colorable ⋮ Every planar graph without 5-cycles and \(K_4^-\) and adjacent 4-cycles is \((2, 0, 0)\)-colorable ⋮ Note on improper coloring of $1$-planar graphs ⋮ Planar graphs without adjacent cycles of length at most five are \((1,1,0)\)-colorable ⋮ \((1,0,0)\)-colorability of planar graphs without cycles of length \(4\) or \(6\) ⋮ A relaxation of Novosibirsk 3-color conjecture

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