Planar graphs without 4-cycles adjacent to triangles are 4-choosable

Cheng, Pengfei; Chen, Min; Wang, Yingqian

doi:10.1016/j.disc.2016.06.009

Cited by 17 publications

(14 citation statements)

References 8 publications

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“…Let C k be the cycle of length k. Lam, Xu, and Liu [12] verified that every planar graph without C 4 is 4-choosable. And Cheng, and Chen, Wang [6], and Kim and Ozeki [11] extended the result independently by certifying the following theorem. In this paper, we extend Theorem 1.3 by proving the following theorem.…”

Section: Introductionmentioning

confidence: 88%

Planar Graphs Without 4-Cycles Adjacent to Triangles are DP-4-Colorable

Kim

2019

Graphs and Combinatorics

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Section: Introductionmentioning

confidence: 88%

Planar Graphs Without 4-Cycles Adjacent to Triangles are DP-4-Colorable

Kim

2019

Graphs and Combinatorics

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“…(1) Suppose G contains C (3,3,4). Then we have four pairwise adjacent cycles x 1 x 2 x 3 , x 1 x 2 x 3 x 4 , x 1 x 3 x 4 x 5 x 6 , and x 1 x 2 x 3 x 4 x 5 x 6 , contrary to G ∈ A.…”

Section: Preliminariesmentioning

confidence: 99%

“…(3) Suppose G contains C (3,4,3). Then we have four pairwise adjacent cycles x 1 x 2 x 3 , x 1 x 3 x 4 x 5 ,…”

Section: Preliminariesmentioning

confidence: 99%

“…More conditions for a planar graph to be 4-choosable are investigated. It is shown that a planar graph is 4-choosable if it has no 4-cycles [10], 5-cycles [15], 6-cycles [7], 7-cycles [6], intersecting 3-cycles [16], intersecting 5-cycles [9], or 3-cycles adjacent to 4-cycles [3,4]. Xu and Wu [14] proved that if every 5-cycle of a planar graph G is not simultaneously adjacent to 3-cycles and 4-cycles, then G is 4-choosable.…”

Section: Introductionmentioning

confidence: 99%

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Planar graphs without pairwise adjacent 3-,4-,5-, and 6-cycle are 4-choosable

Sittitrai¹,

Nakprasit²

2018

Preprint

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Xu and Wu proved that if every 5-cycle of a planar graph G is not simultaneously adjacent to 3-cycles and 4-cycles, then G is 4-choosable. In this paper, we improve this result as follows. If G is a planar graph without pairwise adjacent 3-,4-,5-, and 6−cycle, then G is 4-choosable.x 1 x 2 x 3 x 4 x 5 , and x 1 x 2 x 3 x 4 x 5 x 6 , contrary to G ∈ A.Suppose G contains C(3, 4, 4). Then we have four pairwise adjacent cycles x 1 x 2 x 3 , x 1 x 3 x 4 x 5 , x 1 x 2 x 3 x 4 x 5 , and x 1 x 3 x 4 x 5 x 6 x 7 , contrary to G ∈ A.(4) Suppose G contains C(4, 3, 5). Then we have four pairwise adjacent cycles x 1 x 4 x 5 , x 1 x 2 x 3 x 4 , x 1 x 2 x 3 x 4 x 5 , and x 1 x 4 x 5 x 6 x 7 x 8 , contrary to G ∈ A.(5) Let the hub of W 5 be q and let external vertices be r, s, u, and v in a cyclic order. Suppose there is a cycle uvw. Then we have four pairwise adjacent cycles vwu, vwuq, vwusq, and vwusqr, contrary to G ∈ A.Suppose there is a cycle uvwx. Then we have four pairwise adjacent cycles usq, usqv, usqrv, and usqvwx, contrary to G ∈ A.Suppose there is a cycle uvwxy. Then we have four pairwise adjacent cycles uqv, uqrv, uqsrv, and uqvwxy, contrary to G ∈ A.Suppose there is a cycle uvwxyz. Then we have four pairwise adjacent cycles uvq, uvqs, uvqrs, and uvwxyz, contrary to G ∈ A.Lemma 2.3. If C is a 6-cycle with a triangular chord, then C has exactly one chord.Proof. Let C = tuvxyz with a chord tv. Suppose to the contrary that C has another chord e. By symmetry, it suffices to assume that e = ux, uy, tx, ty, or xz.If e = ux, then we have four pairwise adjacent cycles tuv, tuxv, tvxyz, and tuvxyz, contrary to G ∈ A.If e = uy, then we have four pairwise adjacent cycles tuv, uvxy, tvxyz, and tuvxyz, contrary to G ∈ A.If e = tx, then we have four pairwise adjacent cycles tuv, tuvx, tvxyz, and tuvxyz, contrary to G ∈ A.If e = ty, then we have four pairwise adjacent cycles tuv, tvxy, tvxyz, and tuvxyz, contrary to G ∈ A.If e = xz, then we have four pairwise adjacent cycles tuv, tvxz, tvxyz, and tuvxyz, contrary to G ∈ A.Thus C has exactly one chord. StructureTo prove Theorem 1.1, we prove a stronger result as follows.Theorem 3.1. If G ∈ A with a 4-assignment L, then each precoloring of a 3-cycle in G can be extended to an L-coloring of G.If G does not contain a 3-cycles, then G is 4-choosable as stated above. So we consider (G, C 0 ) and a 4-assignment L where C 0 is a precolored 3-cycle as a minimal counterexample to Theorem 3.1. Embed G in the plane. Lemma 3.2. G has no separating 3-cycles.

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“…Farzad [5] proved that all planar graphs without 7-cycles are 4-choosable. Cheng, Chen and Wang [3] proved that planar graphs without 4-cycles adjacent to triangles are 4-choosable. In the paper, we prove that all planar graphs without intersecting 5-cycles are 4-choosable.…”

Section: Introductionmentioning

confidence: 99%