Planar graphs without cycles of length 4 or 5 are <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" altimg="si2.gif" display="inline" overflow="scroll"><mml:mrow><mml:mo>(</mml:mo><mml:mn>2</mml:mn><mml:mo>,</mml:mo><mml:mn>0</mml:mn><mml:mo>,</mml:mo><mml:mn>0</mml:mn><mml:mo>)</mml:mo></mml:mrow></mml:math>-colorable

Chen, Ming; Wang, Yingqian; Li, Peipei; Xu, Jinghan

doi:10.1016/j.disc.2015.10.029

Cited by 21 publications

(20 citation statements)

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“…So, let us assume f is not a (3, 4 − , 4)-face. By R2, v sends to f charge at most 8 3 , and to abnormal vertices on f a total charge at most 1 6 × 2 when R8 is applicable for v. Moreover, Combining Lemma 3.11 and the rule R6 yields that v sends to possible pendent 3-faces a total charge at most max{ 5 3 + 5 4 , 3 2 × 2}, equal to 3. Therefore, ch * (v) ≥ ch(v) − 8 3 − 1 6 × 2 − 3 = 0.…”

Section: Structural Properties Of the Minimal Counterexample Gmentioning

confidence: 95%

“…R6. Every internal 4 + -vertex sends to each pendent 3-face f charge 5 3 if f is (3, 3, 3)-face, charge 3 2 if f is a (3, 3, 4)-face, and charge 5 4 otherwise. R7.…”

Section: Structural Properties Of the Minimal Counterexample Gmentioning

confidence: 99%

“…Xu [19] showed that planar graphs with neither adjacent triangles nor cycles of length 5 are (1, 1, 1)-colorable. So far, the best known results for planar graphs having no cycles of length 4 or 5 are that, they are (1,1,0)-colorable [10,21] and also (2,0,0)-colorable [5], improving on some results in [9,19]. Because of the refutal of Steinberg's conjecture, the following question is the only one in this direction that remains open.…”

Section: Introductionmentioning

confidence: 99%

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(1,0,0)-colorability of planar graphs without cycles of length 4 or 6

Wang

Yang

2020

Preprint

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A graph G is (d1, d2, d3)-colorable if the vertex set V (G) can be partitioned into three subsets V1, V2 and V3 such that for i ∈ {1, 2, 3}, the induced graph G[Vi] has maximum vertex-degree at most di. So, (0, 0, 0)-colorability is exactly 3-colorability.The well-known Steinberg's conjecture states that every planar graph without cycles of length 4 or 5 is 3-colorable. As this conjecture being disproved by Cohen-Addad etc. in 2017, a similar question, whether every planar graph without cycles of length 4 or i is 3-colorable for a given i ∈ {6, . . . , 9}, is gaining more and more interest. In this paper, we consider this question for the case i = 6 from the viewpoint of improper colorings. More precisely, we prove that every planar graph without cycles of length 4 or 6 is (1,0,0)-colorable, which improves on earlier results that they are (2,0,0)-colorable and also (1,1,0)-colorable, and on the result that planar graphs without cycles of length from 4 to 6 are (1,0,0)-colorable.

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Section: Structural Properties Of the Minimal Counterexample Gmentioning

confidence: 95%

“…R6. Every internal 4 + -vertex sends to each pendent 3-face f charge 5 3 if f is (3, 3, 3)-face, charge 3 2 if f is a (3, 3, 4)-face, and charge 5 4 otherwise. R7.…”

Section: Structural Properties Of the Minimal Counterexample Gmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

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(1,0,0)-colorability of planar graphs without cycles of length 4 or 6

Wang

Yang

2020

Preprint

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“…One way to relax the conjecture is allowing some color classes to be improper. For every planar graph G without 4-cycles and 5-cycles, Xu, Miao, and Wang [17] proved that G is (1, 1, 0)-colorable, and Chen et al [8] proved that G is (2, 0, 0)-colorable.…”

Section: Introductionmentioning

confidence: 99%

Defective 2-colorings of planar graphs without 4-cycles and 5-cycles

Sittitrai

Nakprasit

2018

Discrete Mathematics

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“…It is shown in [13,14,20] that planar graphs without 4-cycles or 5-cycles are (3, 0, 0)and (1, 1, 0)-colorable. Chen et al [7] further proved that such graphs are (2, 0, 0)-colorable.…”

mentioning

confidence: 99%

A relaxation of the strong Bordeaux Conjecture

Huang

2017

Journal of Graph Theory

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Abstract. Let c1, c2, · · · , c k be k non-negative integers. A graph G is (c1, c2, · · · , c k )-colorable if the vertex set can be partitioned into k sets V1, V2, . . . , V k , such that the subgraph G[Vi], induced by Vi, has maximum degree at most ci for i = 1, 2, . . . , k. Let F denote the family of plane graphs with neither adjacent 3-cycles nor 5-cycle. Borodin and Raspaud (2003) conjectured that each graph in F is (0, 0, 0)-colorable. In this paper, we prove that each graph in F is (1, 1, 0)-colorable, which improves the results by Xu (2009) and Liu-Li-Yu (2014+).

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

Cited by 21 publications

References 20 publications

(1,0,0)-colorability of planar graphs without cycles of length 4 or 6

(1,0,0)-colorability of planar graphs without cycles of length 4 or 6

Defective 2-colorings of planar graphs without 4-cycles and 5-cycles

A relaxation of the strong Bordeaux Conjecture

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